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Students can track their progress and improvement through regular use of S Chand ISC Maths Class 11 Solutions Chapter 21 Measures of Dispersion Chapter Test.

S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Chapter Test

Question 1.
Find the mean deviation from the mean for the following data :
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
Mean \(\bar{x}\) = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}\) = \(\frac{500}{10}\) = 50

xi	38	70	48	40	42	55	63	46	54	44
|xi – \(\bar{x}\) |	12	20	2	10	8	5	13	4	4	6	Σ |xi – \(\bar{x}\) | = 84

∴ M.D about mean = \(\frac{\Sigma\left|x_i-\bar{x}\right|}{n}\) = \(\frac{84}{10}\) = 8.4

Question 2.
Find the mean deviation from the mean for the following data:

xi	3	5	7	9	11	13
fi	6	8	15	25	8	4

Solution:

xi	fi	fixi	| xi – \(\bar{x}\) |	fi | xi – \(\bar{x}\) |
3	6	18	5	30
5	8	40	3	24
7	15	105	1	15
9	25	225	1	25
11	8	88	3	24
13	4	52	5	20
	Σxi = 66	Σfixi = 528		Σfi | xi – \(\bar{x}\) | = 138

By direct method, Mean \(\bar{x}\) = \( \frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{528}{66}\) = 8
Thus M.D from mean = \(\frac{\sum f_i\left|x_i-\bar{x}\right|}{\sum f_i}\) = \(\frac{138}{66}\) = 2.09

Question 3.
Find the mean deviation for the mean for the following data:

Classes	0-10	10-20	20-30	30-40	40-50	50-60
Frequencies	6	8	14	16	4	2

Solution:

Classes	Frequencies fi	Mid-Marks xi	fi xi	| xi – 27 |	fi | xi – 27 |
0-10	6	5	30	22	132
10-20	8	15	120	12	96
20-30	14	25	350	2	28
30-40	16	35	560	8	128
40-50	4	45	180	18	72
50-60	2	55	110	28	56
	Σfi = 50		Σfixi = 1350		Σfi | xi – 27| = 512

Thus by direct method, Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{1350}{50}\) = 27
∴ M.D about mean = \(\frac{\Sigma f_i\left|x_i-27\right|}{\Sigma f_i}\) = \(\frac{512}{50}\) = 10.24

Question 4.
Find the mean deviation about the median for the following data :
11, 3, 8, 7, 5, 14, 10, 2, 9
Solution:
Arranging the given data in ascending order ; we have
2, 3, 5, 7, 8, 9, 10, 11, 14
Here no. of observations = n = 9 (odd)
∴ M_d = \(\left(\frac{n+1}{2}\right)\)th observation = \(\left(\frac{9+1}{2}\right)\)th obs = 5th obs = 8

xi	| xi – Md |
2	6
3	5
5	3
7	1
8	0
9	1
10	2
11	3
14	6
	Σ | xi – Md | = 27

∴ M.D about Median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{27}{9}\) = 3

Question 5.
Find the variance and standard deviation of the following data :

xi	92	93	97	98	102	104	109
fi	3	2	3	2	6	3	3

Solution:
The table of values is given as under:

xi	fi	di = xi – A A = 98	\( d_i^2 \)	fidi	\( f_i d_i^2 \)
92	3	-6	36	-18	108
93	2	-5	25	-10	50
97	3	-1	1	-3	3
98	2	0	0	0	0
102	6	4	16	24	96
104	3	6	36	18	108
109	3	11	12	33	363
	Σfi = 22			Σ fidi = 44	Σ\( f_i d_i^2 \) = 728

∴ Variance = \(\frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2\) = \(\frac{728}{22}\) – \(\left(\frac{44}{22}\right)^2\) = 33.090909 – 4 = 29.09
and S.D = \(\sqrt{\text { Variance }}\) = \(\sqrt{29.09}\) = 5.3935

Question 6.
Calculate the mean and variance after the following data :

Classes	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequency (f)	2	3	5	10	3	5	2

Solution:
The table of values is given as under:

Classes	Frequency fi	xi	di = xi – A	Ui = \(\frac{d_i}{i}\)	fiui	\(f_i u_i^2\)
0-30	2	15	-90	-3	-6	18
30-60	3	45	-60	-2	-6	12
60-90	5	75	-30	– 1	-5	5
90-120	10	105	0	0	0	0
120-150	3	135	30	1	3	3
150-180	5	165	60	2	10	20
180-210	2	195	90	3	6	18
	Σfi = 30				Σfiui = 2	Σ\(f_i u_i^2\) = 76

Well-structured S Chand ISC Maths Class 11 Solutions Chapter 21 Measures of Dispersion Ex 21(a) facilitate a deeper understanding of mathematical principles.

S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Ex 21(a)

Question 1.
15, 17, 19, 25, 30, 35, 48
Solution:
The table of values is given as under :

xi	di = xi – \(\bar{x}\)	| di |
15	– 12	12
17	– 10	10
19	-8	8
25	-2	2
30	3	3
35	8	8
48	21	21
Σxi = 189		Σ | di | = 64

Question 2.
21, 23, 25, 28,30, 32, 38, 39, 46, 48
Solution:
Here Mean \(\vec{x}\) = \(\frac{21+23+25+28+30+32+38+39+46+48}{10}\) = \(\frac{330}{10}\) = 3
The table of values is given as under:

xi	xi = xi – \(\bar{x}\)	| di |
21	– 12	12
23	– 10	10
25	-8	8
28	-5	5
30	-3	3
32	– 1	1
38	5	5
39	6	6
46	13	13
48	15	15
		Σ | di | = 78

∴ M.D. about Mean = \(\frac{\Sigma\left|d_i\right|}{n}\) = \(\frac{78}{10}\) = 7.8 and coeff. of M.D. = \(\frac{\text { M.D }}{\bar{x}}\) = \(\frac{7.8}{33}\) = 0.236

Question 3.
10, 70, 50, 53, 20, 95, 55, 42, 60, 48, 80
Calculate the mean deviation from the mean for the following frequency distributions.
Solution:
Mean \(\bar{x}\) = \(\frac{10+70+50+53+20+95+55+42+60+48+80}{11}\) = \(\frac { 583 }{ 11 }\) = 53

xi	10	70	50	53	20	95	55	42	60	48	80
xi – \(\bar{x}\) = di	-43	17	-3	0	-33	42	2	– 11	7	-5	27
| di |	43	17	3	0	33	42	2	11	7	5	27	Σ | di | = 190

∴ required M.D about Mean = \(\frac{\Sigma\left|d_i\right|}{n}\) = \(\frac{190}{11}\) = 17.27 and coeff. of M.D = \(\frac{\text { M.D }}{\text { Mean }}\) = \(\frac{190}{11 \times 53}\) = 0.326

Question 4.

xi	3	9	17	23	27
fi	8	10	12	9	5

Solution:

xi	fi	fixi	| di | = | xi – \(\bar{x}\) |	fi | di |
3	8	24	12	96
9	10	90	6	60
17	12	204	2	24
23	9	207	8	72
27	5	135	16	60
	Σfi = 44	Σ fixi = 660	Σ| di | = 40	Σfi | di | = 312

∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{660}{44}\) = 15
Thus M.D about Mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{312}{44}\) = 7.09

Question 5.

xi	10	11	12	13	14
fi	3	12	18	12	3

Solution:
The table of values is given as under :

xi	fi	di xi	| di | = | xi – \(\bar{x}\) |	fi | di |
10	3	30	2	6
11	12	132	1	12
12	18	216	0	0
13	12	156	1	12
14	3	42	2	6
	Σfi = 48	Σfixi = 576		Σ fi | di | = 36

∴ \(\bar{x}\) = \(\frac { 576 }{ 48 }\) = 12
∴ M.D about Mean = \(\frac{\Sigma f_i\left|d_i\right|}{48}\) = \(\frac{36}{48}\) = \(\frac{3}{4}\) = 0.75

Question 6.

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	5	8	15	16	6

Solution:
We construct the table of values is given as under :

Marks	Class Mark	fi	fixi	| xi – \(\bar{x}\) |	fi | xi – \(\bar{x}\) |
0-10	5	5	25	22	110
10-20	15	8	120	12	96
20-30	25	15	375	2	30
30-40	35	16	560	8	128
40-50	45	6	270	18	108
		Σfi = 5o	Σfixi = 1350		Σ fi | xi – \(\bar{x}\) | = 472

∴ by direct method, mean \(\bar{x}\) = \(\frac{\sum f_i x_i}{\sum f_i}\) = \(\frac{1350}{50}\) = 27
Thus, M.D from mean = \(\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}\) = \(\frac{472}{50}\) = 9.44

Question 7.

Scores	140-150	150-160	160-170	170-180	180-190	190-200
NO. of students	4	6	10	18	9	3

Solution:
The table of values is given as under:

Scores	No. of students fi	xi	fixi	| di | = | xi – \(\bar{x}\) |	fi | di |
140-150	4	145	580	26.2	104.8
150-160	6	155	930	16.2	97.2
160-170	10	165	1650	6.2	62
170-180	18	175	3150	3.8	68.4
180-190	9	185	1665	13.8	124.2
190-200	3	195	585	23.8	71.4
	Σfi = 50		Σfixi = 8560		Σfi | di | = 528

∴ Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{8560}{50}\) = 171.2
Thus M.D about mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{528}{50}\) = 10.56

Question 8.

Class Interval	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	3	50	84	32	10	3

Solution:

Class	Frequency	Mid-Marks	fixi	| di | = | xi – \(\bar{x}\) |	fi | di |
0-20	3	10	30	40.55	121.65
20-40	50	30	1500	20.55	1027.50
40-60	84	50	4200	0.55	46.2
60-80	32	70	2240	19.45	622.40
80-100	10	90	900	39.45	394.5
100- 120	3	110	330	59.45	178.35
	Σfi = 182		Σfixi = 9200		Σfi | di | = 2390.6

∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{9200}{182}\) = 50.55
∴ M.D about mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{2390.6}{182}\) = 13.135

Question 9.
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Solution:
Arranging the data in ascending order, we get 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
Here no. of observations = n = 11 (odd)
∴ \(\mathrm{M}_d=\left(\frac{n+1}{2}\right) \mathrm{th}\) observation = 6th observation = 9

xi	3	3	4	5	7	9	10	12	18	19	21
| xi – Md |	+ 6	6	5	4	2	0	1	3	9	10	12	Σ | xi – Md | = 58

∴ M.D about median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{58}{11}\) = 5.273
and Coefficient of M.D = \(\frac{\text { M.D }}{\text { Median }}\) = \(\frac{58}{11 \times 9}\) = 0.586

Question 10.
100, 150, 200, 250,360, 490, 500, 600, 671
Solution:
Data given is already in ascending order we have, no. of observations = n = 9 (odd)
∴ \(\mathrm{M}_d=\left(\frac{n+1}{2}\right) \mathrm{th}\) observation = 5th observation = 360

xi	100	150	200	250	360	490	500	600	671
| xi – Md |	260	210	160	110	0	130	140	240	311	Σ | xi – Md | = 58

∴ M.D about median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{1561}{9}\) = 173.44
and Coeff. of M.D = \(\frac{\text { M.D }}{\text { median }}\) = \(\frac{1561}{9 \times 360}\) = 0.48

Question 11.

x	10	11	12	13	14
f	3	12	18	12	3

Solution:
The table of values is given as under :

xi	fi	Cumulative freq.	| xi – Md |	fi | xi – Md |
10	3	3	2	6
11	12	15	1	12
12	18	33	0	0
13	12	45	1	12
14	3	48	2	6
	Σ fi = n = 48			Σ fi | xi – Md | = 36

Question 12.

x	3	6	9	12	13	15	21	22
f	3	4	5	2	4	5	4	3

Solution:
The table of values is given as under :

x	f	c.f	| x – Md |	f | x – Md |
3	3	3	10	30
6	4	7	7	28
9	5	12	4	20
12	2	14	1	2
13	4	18	0	0
15	5	23	2	10
21	4	27	8	32
22	3	30	9	27
	Σf = 30			Σf| x – Md |= 149

Accessing S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(f) can be a valuable tool for students seeking extra practice.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(f)

Question 1.
(i) In a single throw of two coins, find the probability of getting both heads or both tails.
(ii) A dice is thrown twice. Find the probability that the sum of the two numbers obtain is 5 or 7 ?
(iii) Two dice are tossed once. Find the probability of getting an even number on first die or a total of 8.
Solution:
(i) In a single throw of two coins
Then sample space S = {HH, HT, TH, TT}
∴ Total no. of exhaustive cases = 4
P (both heads) = \(\frac { 1 }{ 4 }\) {HH}
P (both tails) = \(\frac { 1 }{ 4 }\) {TT}
Thus required probability = P (both heads) + P (tails) = \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

(ii) Here total no. of outcomes = 6² = 36 = n (S)
Let A : sum of two numbers be 5 = {(1, 4), (2, 3), (3, 2), (4, 1)} ∴ n (A) = 4
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{4}{36}\)
B : event that sum of numbers be 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} ∴ n (B) = 6
Thus P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{6}{36}\)
∴ required probability of getting either sum 5 or 7 = P (A or B) = P (A) + P (B) = \(\frac { 4 }{ 36 }\) + \(\frac { 6 }{ 36 }\) = \(\frac { 10 }{ 36 }\) = \(\frac { 5 }{ 18 }\)
[since A and B are mutually exclusive events as A ∩ B = Φ) P (A ∩ B) = 0]

(iii) Here, total no. of exhaustive cases = n (S) = 6² = 36
Let A : event of getting an even number on first die
= {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 18

Question 2.
If the probability of a horse A winning a race is \(\frac { 1 }{ 5 }\) and the probability of horse B winning the same race is \(\frac { 1 }{ 4 }\), what is the probability that one of the horses will win ?
Solution:
Given P (horse A winning a race) = \(\frac { 1 }{ 5 }\); P (horse B winning a race) = \(\frac { 1 }{ 4 }\)
∴ required probability = P (A) + P(B) = \(\frac { 1 }{ 5 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 9 }{ 20 }\)

Question 3.
In a single throw of two dice, what is the probability of obtaining a total of 9 or 11 ?
Solution:
In a single throw of two dice
Then total no. of exhaustive cases = 6² = 36 = n (S)
Let A : event of obtaining a total of 9 = {(3, 6), (4, 5), (5,4), (6, 3)} ∴n (A) = 4
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 26 }\)
Let B : obtaining a total of 11 = {(5, 6), (6, 5)} ∴ n(B) = 2
Thus P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 36 }\)
Here A ∩ B = Φ ⇒ P(A ∩ B) = 0
Thus required probability of obtaining a total of 9 or 11 = P (A or B)
= P (A) + P (B) – P (A ∩ B) = \(\frac { 4 }{ 36 }\) + \(\frac { 2 }{ 36 }\) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Question 4.
In a group there are 2 men and 3 women. 3 persons are selected at random from the group. Find the probability that 1 man and 2 women or 2 men and 1 woman are selected.
Solution:
Total no. of persons = 2 + 3 = 5
∴ Total no. of ways of selecting 3 persons out of 5 = ⁵C₃
Let A : event that 1 man and 2 women are selected ∴ n(A) = ²C₁ × ³C₂

Question 5.
In a class of 25 students with roll numbers 1 to 25, a student is picked up at random to answer a question. Find the probability that the roll number of the selected student is either a multiple of 5 or 7 ?
Solution:
Total exhaustive cases = 25 = n (S)
Let A : event that roll no. of selected student be a multiple of 5 = {5, 10, 15, 20, 25}
B : event that roll no. of selected student be a multiple of 7 = {7, 14,21}
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
∴ P(A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{5}{25}\); P(B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{3}{25}\)
Thus required probability = P(A or B) = P(A) + P(B) = \(\frac { 5 }{ 25 }\) + \(\frac { 3 }{ 25 }\) = \(\frac { 8 }{ 25 }\)

Question 6.
If chance of A, winning a certain race be \(\frac { 1 }{ 6 }\) and the chance of B winning it is \(\frac { 1 }{ 3 }\), what is the chance that neither should win ?
Solution:
Given P(A’s Winning) = \(\frac { 1 }{ 6 }\), P(B’s Winning) = \(\frac { 1 }{ 3 }\)
∴ P (neither A nor B wins)

Question 7.
Discuss and criticise the following : P(A) = \(\frac { 2 }{ 3 }\), P(B) = \(\frac { 1 }{ 4 }\), P (C) = \(\frac { 1 }{ 3 }\) are the probabilities of three mutually exclusive events A, B and C.
Solution:
Since A, B and C are mutually exclusive events.
Now P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = \(\frac { 2 }{ 3 }\) + \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 5 }{ 4 }\) > 1
But P(A ∪ B ∪ C) ≤ 1 Thus, given statement is false.

Question 8.
E and F are two events associated with a random experiment for which P (F) = 0.35, P (E or F) = 0.85, P (E and F) = 0.15. Find P (E).
Solution;
Given P (F) = 0.35 ; P (E or F) = 0.85 and P (E and F) = 0.15
We know that P (E ∪ F) = P (E) + P (F) – P (E ∩ F)
⇒ 0.85 = P(E) + 0.35 – 0.15
⇒ P (E) = 0.85 – 0.20 = 0.65

Question 9.
(i) Two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Find the probability that neither A nor B occurs.
(ii) The probability of an event A occurring is 0.5 and of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B occurring.
Solution:
(i) Given P (A) = 0.25 ; P (B) = 0.50 and P (A ∩ B) = 0.14
P (neither A nor B) = P(\(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\)) = P \(P(\overline{A \cup B})\) = 1 – P(A ∪ B)
= 1 – [P (A) + P (B) – P (A ∩ B)] = 1 – [0.25 + 0.50 – 0.14] = 1 – 0.61 = 0.39

(ii) Given P (A) = 0.5 ; P (B) = 0.3
Since A and B are mutually exclusive events. ∴ A ∩ B = Φ ⇒ P (A ∩ B) = 0
P (neither A nor B) = P(\(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}\)) = P \(P(\overline{A \cup B})\) = 1 – P(A ∪ B)
= 1 – [P (A) + P (B) – P (A ∩ B)] = 1 – 0.5 – 0.3 – 0 = 0.2

Question 10.
(i) A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is event ?
(ii) A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Solution:
(i) Total no. of ways of drawing 2 tickets from 25 tickets = ²⁵C₂
The product of numbers is even when either both are even or one is even and the other is odd
Let A : event that both are even and there are 12 even numbers from 1 to 25
∴ n (A) = Total no. of ways of drawing 2 tickets out of 12 = ¹²C₂
Let B : event that one is even and other is odd since there are 12 even and 13 odd numbers from 1 to 25.

(ii) Total no. of balls = 7 + 5 + 4 = 16
n (S) = Total no. of ways of drawing four balls out of 16 without replacement = ¹⁶C₄
Drawing atleast three black balls means drawing 3 black balls out of 5 and one ball from remaining 11 balls or 4 black balls out of 5.
Thus required probability = \(\frac{{ }^5 C_3 \times{ }^{11} C_1+{ }^5 C_4}{{ }^{16} C_4}\) = \(\frac{\frac{5 \times 4}{2} \times 11+5}{\frac{16 \times 15 \times 14 \times 13}{24}}\) = \(\frac { 23 }{ 364 }\)

Question 11.
(i) A and B are two mutually exclusive events of an experiment.
If P(not A) = 0.75, P(A ∪ B) = 0.65 and P(B) = p, find the value of p.
(ii) A and B are two mutually exclusive events. If P (A) = 0.5 and P \((\overline{\mathbf{B}})\) = 0.6, find P (A or B).
Solution:
(i) Given A and B are mutually exclusive events.
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
Given P (not A) = 0.65 ⇒ 1 – P (A) = 0.65 ⇒ P (A) = 0.35
P(A ∪ B) = 0.65 and P (B) =p
We know that P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 0.65 = 0.35 + p – 0 ⇒ p = 0.30

(ii) Given A and B are two mutually exclusive events.
∴ A ∩ B = Φ ⇒ P(A ∩ B) = 0
Given P (A) = 0.5 ; P\((\overline{\mathbf{B}})\) = 0.6 ⇒ 1 – P (B) = 0.6 ⇒ P (B) = 0.4
∴ P (A or B) = P (A) + P (B) – P (A ∩ B) = 0.5 + 0.4 – 0 = 0.9

Question 12.
(i) A, B and C are three mutually exclusive and exhaustive events associated with a random experiment. Find P (A) given that P (B) = \(\frac { 3 }{ 2 }\) P (A) and P (C) = \(\frac { 1 }{ 2 }\) P (B).
Solution:
(i) Given A, B and C are mutually exclusive, exhaustive events
∴ P(A) + P(B) + P(C) = 1 …(1)
Given P(B) = \(\frac { 3 }{ 2 }\) P(A) and P (C) =\(\frac { 1 }{ 2 }\) P(B) = \(\frac { 1 }{ 2 }\) × \(\frac { 3 }{ 2 }\)P(A) = \(\frac { 3 }{ 4 }\) P (A)
∴ from (1); P(A) + \(\frac { 3 }{ 2 }\)P (A) + \(\frac { 3 }{ 4 }\) P(A) = 1
\(\left[1+\frac{3}{2}+\frac{3}{4}\right] \mathrm{P}(\mathrm{A})=1\)
\(\Rightarrow\left(\frac{4+6+3}{4}\right) P(A)=1\)
⇒ P (A) = \(\frac { 4 }{ 13 }\)

(ii) Since A, B and C are mutually exclusive and exhaustive events.
∴ P (A) + P (B) + P (C) = 1
Given P (A) = 2 P (B) = 3 P (C) …(1)

Question 13.
(i) In a single throw of two dice, find the probability that neither a doublet nor a total of a 10 will appear.
(ii) Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4.
(iii) Two dice are thrown together ; what is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 5.
Solution:
(i) In a single throw of two dice, total exhaustive cases = 6² = 36
Let A: getting a doublet = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} ∴ n (A) = 6
B : getting a total of 10 = {(4, 6), (5, 5), (6, 4)} ∴n (B) = 3
Thus P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 6 }{ 36 }\); P (B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 36 }\)
and A ∩ B = {(5, 5)} ∴ n (A ∩ B) = 1
Thus P (A ∩ B) = \(\frac{n(\mathrm{~A} \cup \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 36 }\)
∴ P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac { 6 }{ 36 }\) + \(\frac { 3 }{ 36 }\) – \(\frac { 1 }{ 36 }\) = \(\frac { 8 }{ 36 }\) = \(\frac { 2 }{ 9 }\)
Thus, required probability = P (neither A nor B) = \(\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})\) = \(P(\overline{A \cup B})\)

(ii) Total exhaustive cases = n (S) = 6² = 36
Let A : event that the sum of the numbers obtained on two dice is multiple of 3
= {(U 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (A)= 12
Let B = event that sum of the numbers on two dice is multiple of 4
= {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)} ∴ n (B) = 9
∴ A ∩ B = {(6, 6)} ⇒ n (A ∩ B) = 1
Thus, P (A) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac{12}{36}\); P (B) = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{9}{36}\) and P (A ∩ B) = \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 36 }\)
∴ P(A ∪ B) = P (A) + P (B) = P (A ∩ B) = \(\frac { 12 }{ 36 }\) + \(\frac { 9 }{ 36 }\) – \(\frac { 1 }{ 36 }\) = \(\frac { 20 }{ 36 }\) = \(\frac { 5 }{ 9 }\)
∴ required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A u B) = 1 – \(\frac { 5 }{ 9 }\) = \(\frac { 4 }{ 9 }\)

(iii) When two dice are thrown together
Then total no. of exhaustive cases = n (S) = 6² = 36
Let A : event that the sum of the numbers on the two faces is divisible by 3
= {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (A) = 12
Let B : event that the sum of numbers on two faces is divisible by 5
= {(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)} ∴ n (B) = 7
and A ∩ B = Φ ⇒ P (A ∩ B) = 0
Thus, P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – 0 = \(\frac { 12 }{ 36 }\) + \(\frac { 7 }{ 36 }\) = \(\frac { 19 }{ 36 }\)
∴ Required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A ∪ B) = 1 – \(\frac { 19 }{ 36 }\) = \(\frac { 17 }{ 36 }\)

Question 14.
In a given race, the odds in favour of horses A, B, C and D are 1 : 3, 1 : 4, 1 : 5 and 1 : 6 respectively. Find the probability that one of them wins the race.
Solution:
Given odds in favour of horse A be 1 : 3
∴ P (horse A’s winning) = \(\frac{1}{1+3}\) = \(\frac{1}{4}\)
odds in favour of horse B be 1 : 4
∴ P (horse B’s winning) = \(\frac{1}{1+4}\) = \(\frac{1}{5}\)
odds in favour of horse C be 1 : 5
∴ P (horse C’s winning) = \(\frac{1}{1+5}\) = \(\frac{1}{6}\)
and odds in favour of horse D be 1 : 6
∴ D (horse D’s winning) = \(\frac{1}{1+6}\) = \(\frac{1}{7}\)
∴ Required probability that one of them will win the race = P (A) + P (B) + P (C) + P (D)
[since A, B, C and D are mutually exclusive and exhaustive events]
= \(\frac{1}{4}\) + \(\frac{1}{5}\) + \(\frac{1}{6}\) + \(\frac{1}{7}\) = \(\frac{105+84+70+60}{420}\) = \(\frac{319}{420}\)

Question 15.
100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both examinations.
Solution:
Let A : event that students pass the fist examination
B : event that students pass the second exam.
Then P(A) = \(\frac{60}{100}\); P (B) = \(\frac{5}{100}\); P(A ∩ B) = \(\frac{60}{100}\) + \(\frac{50}{100}\) – \(\frac{30}{100}\) = \(\frac{80}{100}\)
Thus required probability that selected student failed in both exams = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\)
= 1 – P (A ∪ B) = 1 – \(\frac{80}{100}\) = \(\frac{20}{100}\) = 0.2

Question 16.
(i) A card is drawn from a deck of 52 can is. Find the probability of getting an ace or a spade card.
(ii) From a well shuffled deck of 52 cards, 4 cards are drawn at random. What is the probability that all the drawn cards are of the same colour.
Solution:
(i) Total no. of exhaustive cases = 52
Let A : event of getting an ace card ∴ n (A) = 4
B : getting a spade card ∴ n (B) = 13
A ∩ B : getting an ace of spade
∴ n (A ∩ B) = 1
Thus P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\)
= \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) n(S) = Total no. of ways of drawing 4 cards out of 52 = ⁵²C₄
All drawing cards are of same colour means all cards are either red or black.
Let A : drawing four red cards.
∴ n (A) = Total no. of ways of drawing four cards out of 26 = ²⁶C₄
Let B : drawing four black cards
∴ n (B) = Total no. of ways of drawing four cards out of 26 = ²⁶C₄
Thus required probability =P (A) + P (B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac{2 \times{ }^{26} \mathrm{C}_4}{{ }^{52} \mathrm{C}_4}\)
= 2 × \(\frac{26 \times 25 \times 24 \times 23}{52 \times 51 \times 50 \times 49}\) = \(\frac{92}{833}\)

Question 17.
(i) A card is drawn at random from a well shuffled pack of cards. What is the probability that it is a heart or a queen ?
(ii) A card is drawn at random from a well shuffled pack of 52 cards. Find the probability it is neither a king nor a heart ?
Solution:
(i) Given total exhaustive cases = 52 = n (S)
Let A : event that heart card is drawing ∴ n (A) = 13
and B : drawing a queen card ∴ n (B) = 4
A ∩ B : drawing a queen of heart ∴ n (A ∩ B) = 1
Thus required probability = P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
= \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{13}{52}\) + \(\frac{4}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) Total no. of exhaustive cards = n (S) = 52
Let A : drawing a king card ∴ n (A) = 4
and B : drawing a heart card ∴ n (B) = 13
A ∩ B : drawing a king card of heart ∴ n (A ∩ B) = 1
Thus P (A ∪ B) = P (A) + P (B) – P (A ∩ B) = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) + \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) – \(\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{4}{52}\) + \(\frac{13}{52}\) – \(\frac{1}{52}\) = \(\frac{16}{52}\) = \(\frac{4}{13}\)
∴ Required probability = \(P(\bar{A} \cap \bar{B})\) = \(P(\overline{A \cup B})\) = 1 – P (A ∪ B) = 1 – \(\frac{4}{13}\) = \(\frac{9}{13}\)

Question 18.
A card is drawn from a pack of 52 cars. Find the probability of getting a king or a heart or a red card.
Solution:
Total no. of outcomes = n (S) = 52
Let A : drawing a king card ∴ n (A) = 4
B : drawing a heart card ∴ n (B) = 13
C : drawing a red card ∴ n (C) = 26
A ∩ B : drawing a king of heart card
∴ n( A ∩ B) = 1
B ∩ C : drawing a heart card
∴ n( B ∩ C) = 13
A ∩ C : drawing a king of red card
∴ n( A ∩ C) = 2
∴ A ∩ B ∩ C : drawing a king of heart card
∴ n(A ∩ B ∩ C) = 1
Required probability = P(A ∪ B ∪ C) = P (A) + P (B) + P(C) – P (A ∩ B) -P (B ∩ C) – P (A ∩ C) + P(A ∩ B ∩ C)
= \(\frac{4}{52}\) + \(\frac{13}{52}\) + \(\frac{26}{52}\) –\(\frac{1}{52}\) – \(\frac{13}{52}\) – \(\frac{2}{52}\) + \(\frac{1}{52}\) = \(\frac{28}{52}\) = \(\frac{7}{13}\)

Question 19.
There are three events A, B, C one of which must and only one can happen. The odds are 8 to 3 against A, 5 to 2 against B ; find the odds against C.
Solution:
Given odds against A be 8 : 3.
∴ P (A) = \(\frac{3}{8+3}\) + \(\frac{3}{11}\)
Also, odds against B be 5 : 2
Since out of three events A, B and C, one of which must and only one can happen ∴ A, B and C are mutually exclusive and exhaustive events.
∴ P (A) + P (B) + P (C) = 1
⇒ P (C) = 1 – \(\frac{3}{11}\) – \(\frac{2}{7}\) = \(\frac{77-21-22}{77}\) = \(\frac{34}{77}\)
\(\mathrm{P}(\overline{\mathrm{C}})\) = 1 – P (C) = 1 – \(\frac{34}{77}\) = \(\frac{43}{77}\)
Thus required odds against C be \(\mathrm{P}(\overline{\mathrm{C}})\) : P (C) i.e. \(\frac{43}{77}\) : \(\frac{34}{77}\) i.e. 43 : 34.

Question 20.
In a group of students, there are 3 boys and 3 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution:
Given total no. of students = 3 + 3 = 6
∴ n (S) = Total no. of ways of selecting 4 students out of 6 = ⁶C₄
Let A : selecting 3 boys and 1 girl
∴ n (A) = Total no. of ways of selecting 3 boys out of 3 and 1 girl out of 3 = ³C₃ × ³C₁
Let B : selecting 1 boy and 3 girls
∴ n (B) = Total no. of ways of selecting 1 boy out of 3 and 3 girls out of 3 = ³C₁ × ³C₃
Thus required probability = P (A ∪ B) = P
(A) + P (B) = \(\frac{{ }^3 \mathrm{C}_3 \times{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_3}{{ }^6 \mathrm{C}_4}\) = \(\frac{1 \times 3+3 \times 1}{\frac{6 \times 5}{2}}\) = \(\frac{6}{15}\) =  \(\frac{2}{5}\)

Students often turn to S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(e) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(e)

Question 1.
Sameer throws an ordinary die. What is the probability that he throws
(i) 2
(ii) 5
(iii) 2 or 5 ?
Solution:
(i) When a die is thrown then total no. of outcomes = n (S) = 6 where S = {1, 2,3, 4, 5, 6}
∴ P (getting 2) = \(\frac { 1 }{ 6 }\)
(ii) P (getting 5) = \(\frac { 1 }{ 6 }\)
(iii) P (getting 2 or 5) = P (2) + P (5) = \(\frac { 1 }{ 6 }\) + \(\frac { 1 }{ 6 }\) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

Question 2.
Kavita draws a card from a pack of cards. What is the probability that she draws
(i) a heart
(ii) a club
(iii) a heart or a club ?
Solution:
Total no. of cards = n (S) = 52
(i) since there are 13 heart cards ∴ P (heart) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(ii) p (a club) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(iii) P (a heart or a club) = P (heart) + P (club)  = \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

Question 3.
Anurag draws a card from a pack of cards. What is the probability that he draws one of the following numbers ?
(i) 3
(ii) 7
(iii) 3 or 7
Solution:
Here n (S) = 52
(i) Since there are four 3’s (one each from heart, diamond, club and spade)
∴ required probability of getting 3 = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(ii) P (getting 7) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(iii) P (3 or 7) = P (3) + P (7) = \(\frac { 1 }{ 13 }\) = \(\frac { 1 }{ 13 }\) = \(\frac { 2 }{ 13 }\)

Question 4.
A letter is chosen at random from the letters in the word PROBABILITY. What is the probability that the letter will be
(i) B
(ii) a vowel
(iii) B or a vowel ?
Solution:
Total no. of letters in word PROBABILITY = 11
(i) P (getting a letter B) = \(\frac { 2 }{ 11 }\) [since there are 2B’s]

(ii) since there are 4 vowels O, A, I, I
∴ P (getting a vowel) = \(\frac { 4 }{ 11 }\)

(iii) P (B or a vowel) = P (B) + P (a vowel) = \(\frac { 2 }{ 11 }\) + \(\frac { 4 }{ 11 }\) = \(\frac { 6 }{ 11 }\)

Question 5.
A bag contains 7 white balls, 9 green balls and 10 yellow balls. A ball is drawn at random from the bag. What is the probability that it will be
(i) white
(ii) green
(iii) green or white
(iv) not yellow
(v) neither yellow nor green ?
Solution:
Given no. of white balls = 7, no. of green balls = 9 and no. of yellow balls = 10
∴ Total no. of balls = 7 + 9 + 10 = 26

(i) P(white ball) = \(\frac { 7 }{ 26 }\)

(ii) P(green ball) = \(\frac { 9 }{ 26 }\)

(iii) P(green or white) = P(green) + P(white) = \(\frac { 9 }{ 26 }\) + \(\frac { 7 }{ 26 }\) = \(\frac { 16 }{ 26 }\) = \(\frac { 8 }{ 13 }\)

(iv) P(yellow ball) = \(\frac { 10 }{ 26 }\) ∴ P(getting not yellow ball) = 1 – \(\frac { 10 }{ 26 }\) = \(\frac { 16 }{ 26 }\) = \(\frac { 8 }{ 13 }\)
Thus, P (neither yellow nor green) = P (white ball) = \(\frac { 7 }{ 26 }\)

Question 6.
Suyash needs his calculator for his mathematics lesson. It is either in his pocket, bag or locker. The probability it is in his pocket is 0.20, the probability it is in his bag is 0.58. What is the probability that (i) he will have the calculator for the lesson, (ii) it is in his locker ?

Solution:
(i) Given P (calculator is in his pocket) = 0.20
P (calculator is in his bag) = 0.58
∴ required probability = P (calculator in his pocket) + P (calculator in his bag) = 0.20 + 0.58 = 0.78
(iii) required probability that calculator is in his locker = 1 – P (calculator is in his pocket) – P (calculator is in his bag) = 1 – 0.20 – 0.58 = 0.22

Question 7.
A spinner has numbers and colours on it, as shown in the diagram. Their probabilities are given in the tables. When the spinner is spun what is the probability of each of the following ?

(i) red or green
(ii) 2 or 3
(iii) 3 or green
(iv) 2 or green
(v) Explain why the answer to P (1 or red) is not 0.9.
Solution:
Given P (red) = 0.5 ; P (1) = 0.4
P (green) = 0.25 ; P (2) = 0.35
P (blue) = 0.25 ; P (3) = 0.25
(i) P (red or green) = P (red) + P (green) = 0.5 + 0.25 = 0.75
(ii) P (2 or 3) = P (2) + P (3) = 0.35 + 0.25 = 0.6
(iii) P (3 or green) = P (3) + P (green) = 0.25 + 0.25 = 0.50
(iv) P (2 or green) = P (2) + P (green) = 0.35 + 0.25 = 0.60
(v) Here 1 and red are not mutually exclusive event
∴ P (1 and red) ≠ 0 [∵ 1 ∩ red ≠ Φ]
∴ P (1 or red) ≠ P(1) + P(red) = 0.9

The availability of step-by-step S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(d) can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(d)

Question 1.
Four digit numbers are formed by using the digits 1, 2, 3, 4 and 5 without repeating the digit. Find the probability that a number, chose at random, is an odd number.
Solution:
Total no. of 4 digit numbers formed by using given digits 1, 2, 3, 4 and 5 without repetition

∴ Total no. of exhaustive cases = 120
Total no. of favourable cases = total no. of 4 digit odd number = 2 × 3 × 4 × 3 = 72

Question 2.
What is the probability of getting 3 white balls in a draw of 3 balls from a box containing 6 white and 4 red balls ?
Solution:
Given Total no. of balls = 6 + 4 = 10
Total no. of cases = Total no. of drawing 3 balls out of 10 balls = ¹⁰C₃
Favourable no. of cases = Total no. of ways of drawing 3 white balls out of 6 = ⁶C₃

Question 3.
A bag contains 6 white, 7 red and 5 black balls, three balls are drawn at random. Find the probability that they will be white.
Solution:
Total no. of balls = 6 + 7 + 5 = 18
∴ Total no. of exhaustive cases = Total no. of ways of drawing 3 balls out of 18 = ¹⁸C₃
Total no. of favourable cases = no. of ways of drawing 3 white balls out of 6 = ⁶C₃

Question 4.
A bag contains 2 white marbles, 4 blue marbles, and 6 red marbles. A marble is drawn at random from the bag. What is the probability’ that
(i) it is white ?
(ii) it is blue ?
(iii) it is red ?
(iv) it is not white ?
(v) it is not blue ?
(vi) it is black ?
Solution:
Total number of marbles = 6 + 4 + 2=12
∴ total no. of Exhaustive cases = 12
(i) Required probability of drawing a white ball = \(\frac { 2 }{ 12 }\) = \(\frac { 1 }{ 6 }\)

(ii) Required probability of drawing a blue ball = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)

(iii) Required probability of drawing a red ball = \(\frac { 6 }{ 12 }\) = \(\frac { 1 }{ 2 }\)

(iv) Required probability of drawing not a white ball = probability of drawing either red or blue ball = \(\frac { 6+4 }{ 12 }\) = \(\frac { 10 }{ 12 }\) = \(\frac { 5 }{ 6 }\)

(v) drawing ball is not blue ∴ it is either white or red
∴ Total no. of favourable cases = 2 + 6 = 8
Thus required probability = \(\frac { 8 }{ 12 }\) = \(\frac { 2 }{ 3 }\)

(vi) Since there is no black ball
∴ Total no. of favourable cases = 0
Thus, required probability = \(\frac { 0 }{ 12 }\) = 0

Question 5.
In Q. 4, three marbles are drawn from the bag. What is the probability that
(i) they are all blue ?
(ii) they are all red ?
(iii) they are all white ?
(iv) none of them is red ?
(v) not all of them are red ?
(vi) none is black ?
Solution:
Here total no. of exhaustive cases = Total no. of ways of drawing 3 balls out of 12 = ¹²C₃
(i) Total no. of favourable cases = Total no. of ways of drawing 3 out of 4 = ⁴C₃

(iii) Since we have to draw 3 white marbles but we have only 2 white marbles.
∴ Total no. of favourable cses = 0
Thus required probability = \(\frac{0}{{ }^{12} C_3}\) = 0

(iv) Since none of drawing marble be red
∴Total no. of favourable outcomes = Total no. of ways of drawing 3 marbles out of remaining 6 marbles = ⁶C₃

(v) Required probability = 1 – P(all are red) = 1 – \(\frac { 1 }{ 11 }\) = \(\frac { 10 }{ 11 }\)
(vi) Since there is no black ball
∴ required probability that none is black = probability of drawing ball of any colour out of red, blue or white = 1

Question 6.
Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls, one by one without replacement. What is the probability that (i) both balls are of the same colour, (ii) at least one ball is red ?
Solution:
Given no. of white balls = 2, no. of red balls = 3 and no. of black balls = 4
∴ Total no. of balls = 2 + 3 + 4 = 9
(i) Required probability = P (WW) + P (RR) + P (BB)
= \(\frac { 2 }{ 9 }\) × \(\frac { 1 }{ 8 }\) + \(\frac { 3 }{ 9 }\) × \(\frac { 2 }{ 8 }\) + \(\frac { 4 }{ 9 }\) × \(\frac { 3 }{ 8 }\) = \(\frac { 20 }{ 72 }\) = \(\frac { 5 }{ 18 }\)

(ii) Required probability of drawing atleast one ball is red = P (RW, RB, RR, WR, BR) = P (RW) + P (RB) + P (RR) + P (WR) + P (BR)
= \(\frac { 3 }{ 9 }\) × \(\frac { 2 }{ 8 }\) + \(\frac { 3 }{ 9 }\) × \(\frac { 4 }{ 8 }\) + \(\frac { 3 }{ 9 }\) × \(\frac { 2 }{ 8 }\) + \(\frac { 2 }{ 9 }\) × \(\frac { 3 }{ 8 }\) + \(\frac { 4 }{ 9 }\) × \(\frac { 3 }{ 8 }\) = \(\frac { 42 }{ 72 }\) = \(\frac { 7 }{ 12 }\)

Question 7.
From a pack of cards, three are drawn at random. Find the chance that they are a king, a queen and a knave.
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing three cards out of 52 = ⁵²C₃
Total no. of favourable cases = Total no. of ways of drawing a king, a queen and a knave = ⁴C₁ × ⁴C₁ × ⁴C₁

Question 8.
(i) Two cards are drawn at random from 8 cards numbered from 1 to 8. What is the probability that the sum of the numbers is odd, if the two cards are drawn together ?
(ii) Two cards are drawn at random from a well shuffled pack of 52 cards, show that the chances of drawing two aces is \(\frac { 1 }{ 221 }\).
(iii) From a pack of 52 cards, 3 cards are drawn ‘at random’. Find the probability of drawing exactly two aces.
(iv) Three cards are drawn at a time at random from a well shuffled pack of 52 cards. Find the probability that all the three cards have the same number.
(v) Two cards are drawn from a well shuffled pack of cards without replacement. Find the probability that neither a Jack nor a card of spades is drawn.
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing two cards out of 8 = ⁸C₂ = n (S)

(i) Let E : the sum of the numbers is odd
= {(1, 2), (2, 1), (1, 4), (4, 1), (1, 6), (6, 1), (1, 8), (8, 1), (2, 3), (3, 2), (2, 5), (5, 2), (2, 7), (7, 2), (3, 4), (4, 3), (3, 6), (6, 3), (3, 8), (8, 3), (4, 5), (4, 7), (5, 4), (7, 4), (6, 5), (5, 6), (5, 8), (8, 5), (6, 7), (7, 6), (8, 7), (7, 8)}
Since cards are taken together
∴ (a, b) = (b, a)
Thus n (E) = 16
∴ required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac{16}{{ }^8 C_2}\) = \(\frac{16}{28}\) = \(\frac{4}{7}\)

(ii) Total no. of exhaustive cases = Total no. of ways of drawing 2 cards out of 52 = ⁵²C₂
Total no. of favourable outcomes = Total no. of ways of drawing 2 out of 4 = ⁴C₂
∴ required probability = \(\frac{{ }^4 C_2}{{ }^{52} C_2}\) = \(\frac{\frac{4 !}{2 ! 2 !}}{\frac{52 !}{50 ! 2 !}}\) = \(\frac{6}{26 \times 51}\) = \(\frac{1}{13 \times 17}\) = \(\frac { 1 }{ 221 }\)

(iii) Total no. of ways of drawing 3 cards out of 52 = ⁵²C₃
∴ Total no. of favourable outcomes = Total no. of ways of drawing exactly two aces = ⁴C₂ × ⁴⁸C₁

(iv) Total no. of exhaustive case = total no. of ways of drawing 3 cards out of 52 = ⁵²C₃
Let E : all the three cards are having the same number.
Since there are 13 such groups each containing 4 same cards
∴ n (E) = 13 × ⁴C₃ = 13 × 4 = 52

(v) Total no. of exhaustive cases = Total no. of ways of drawing 2 cards out of 52 without replacement = ⁵²C₂ = n (S)
E : drawing card is neither a Jack nor a card of spades and there are (52 – 13 – 3) = 36 such cards
∴ n (E) = Total no. of ways of drawing 2 cards out of 36 = ³⁶C₂

Question 9.
Three cards are drawn from a deck of 52 cards. What is the probability that
(i) they are all spades ?
(ii) they are all red cards ?
(iii) none of them is a club ?
(iv) all of them are aces ?
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing 3 cards out of 52 = ⁵²C₃
(i) Since there are 13 spade cards.
Then total no. of favourable cases = ¹³C₃
∴ required Probability = \(\frac{{ }^{13} C_3}{{ }^{52} C_2}\) = \(\frac{\frac{13 !}{10 ! 3 !}}{\frac{52 !}{49 ! 3 !}}\) = \(\frac{13 \times 12 \times 11}{52 \times 51 \times 50}\) = \(\frac{11}{850}\)

(ii) Since there are 26 red cards
∴ Total no. of ways of drawing 3 cards out of 26 = ²⁶C₃
∴ required probability = \(\frac{{ }^{26} C_3}{{ }^{52} C_3}\) = \(\frac{\frac{26 !}{3 ! 23 !}}{\frac{52 !}{49 ! 3 !}}\) = \(\frac{26 \times 26 \times 24}{52 \times 51 \times 50}\) = \(\frac{2}{17}\)

(iii) Since none of the drawing card is club card so the drawing cards can be out of (52 – 13) = 39 cards
∴ Total no. of favourable cases = No. of ways of drawing 3 cards out of 39 = ³⁹C₃

Question 10.
In Q. 7 what are the odds that
(i) all three cards are spades?
(ii) they are all red cards?
(iii) none is a club?
(iv) all of them are aces?
Solution:
Total no. of exhaustive outcomes = Total no. of ways of drawing 3 cards out of 52 = ⁵²C₃

(i) Since there are 13 spade cards
∴ Total no. of ways of drawing 3 cards out of 13 = ¹³C₃

(ii) Total no. of favourable cases = Total no. of ways of drawing 3 cards out of 26 = ²⁶C₃
∴ required probability = \(\frac{{ }^{26} C_3}{{ }^{52} C_3}\) = \(\frac{26 \times 25 \times 24}{52 \times 51 \times 50}\) = \(\frac{2}{17}\)
Thus required odds = \(\frac { 2 }{ 17 }\) : 1 – \(\frac { 2 }{ 17 }\) = \(\frac { 2 }{ 17 }\) : \(\frac { 15 }{ 17 }\) = \(\frac { 2 }{ 15 }\)

(iii) Total no. of favourable cases = Total no. of ways drawing 3 cards out of (52 – 13) = 39 cards = ³⁹C₃
∴ required probability of drawing none is a club card out of 3 = \(\frac{{ }^{39} C_3}{{ }^{52} C_3}\) = \(\frac{39 \times 38 \times 37}{52 \times 51 \times 50}\) = \(\frac{703}{1700}\)
Thus required odd = \(\frac{703}{1700}\) : 1 – \(\frac{703}{1700}\) = \(\frac{703}{1700}\) : \(\frac{997}{1700}\) = 703 : 997

(iv) Total no. of ways of drawing 3 ace cards out of 4 = ⁴C₃ = 4
∴ required probability = \(\frac{{ }^4 C_3}{{ }^{52} C_3}\) = \(\frac{4}{\frac{52 \times 51 \times 50}{6}}\) = \(\frac{1}{5525}\)
Thus required odds in favour of favourable event = \(\frac{1}{5525}\) : 1 – \(\frac{1}{5525}\) = 1 : 5524

Question 11.
One card is drawn from a pack of 52 cards, each of the 52 cards being equally like to be drawn. Find the probability that the card drawn is (i) an ace, (ii) red, (iii) either red or king, (iv) red and a king.
Solution:
Total no. of exhaustive cases = 52
(i) since there are 4 ace cards
∴ required probability = \(\frac{4}{52}\) = \(\frac{1}{13}\)

(ii) since there are 26 red cards
∴ required probability = \(\frac{26}{52}\) = \(\frac{1}{2}\)

(iii) Since there are 26 red cards that includes 2 red kings and 2 black kings
∴ Total no. of such cards = 26 + 2 = 28
∴ required probability = \(\frac{28}{52}\) = \(\frac{7}{13}\)

(iv) Since there are 2 red kings in 52 cards
Thus required probability = \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 12.
Four cards are drawn at random from a pack of 52 playing cards, Find the probability of getting
(i) all the four cards of the same suit;
(ii) all the four cards of the same number;
(iii) one card from each suit;
(iv) two red cards and two black cards ;
(v) all cards of the same colour ;
(vi) all face cards ; (King, Queen, Jack)
(vii) four honours of the same suit.
Solution:
Total no. of exhaustive cases = Total no. of ways of drawing 3 cards out of 52 = ⁵²C₄
(i) Total no. of favourable cases = Total no. of ways of drawing 4 cards from same suit = 4 × ¹³C₄ [since there are 4 suits and each suit containing 13 cards]

(ii) Let E : getting all the four cards of the same number
∴ n (E) = 13 × ⁴C₄ = 13
[There are 13 such groups and each group containing 4 same cards each one from each suit]
Thus required probability = \(\frac{13}{{ }^{52} C_4}\) = \(\frac{\frac{13}{52 \times 51 \times 50 \times 49}}{24}\) = \(\frac{13}{270725}\)

(iii) Total no. of ways of drawing one card from each suit = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ [since each suit contains 13 cards each]
∴ required probability = \(\frac { favourable cases }{ total cases }\) = \(\frac{{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1 \times{ }^{13} \mathrm{C}_1}{{ }^{52} \mathrm{C}_4}\) =\(\frac{13 \times 13 \times 13 \times 13}{\frac{52 \times 51 \times 50 \times 49}{24}}\) = \(\frac{2197}{20825}\)

(iv) Total no. of favourable cases = Total no. of ways of drawing 2 red and 2 black cards out of 26 cards each = ²⁶C₂ × ²⁶C₂
Thus required probability = \(\frac{{ }^{26} C_2 \times{ }^{26} C_2}{{ }^{52} C_4}\) = \(\frac{\frac{26 \times 25}{2} \times \frac{26 \times 25}{2}}{\frac{52 \times 51 \times 50 \times 49}{24}}\) = \(\frac{325}{833}\)

(v) drawing 4 cards of same colour means either all 4 cards are red or black.
∴ Total favourable cases = Total no. of ways of drawing all four cards of same colour = ²⁶C₄ × ²⁶C₄
∴ required probability = \(\frac{{ }^{26} C_4 \times{ }^{26} C_4}{{ }^{52} C_4}\)

(vi) since there are 12 face cards (4 kings, 4 queens and 4 jacks)
Thus total no. of ways of drawing 4 cards from 12 cards = ¹²C₄
Therefore required probability = \(\frac{{ }^{12} \mathrm{C}_4}{{ }^{52} \mathrm{C}_4}\) = \(\frac{\frac{12 !}{4 ! 8 !}}{\frac{52 !}{4 ! 48 !}}\) = \(\frac{12 \times 11 \times 10 \times 9}{52 \times 51 \times 50 \times 49}\) = \(\frac{99}{54145}\)

(vii) since there are 16 honours cards including 4 aces, 4 kings, 4 queens and 4 jacks i.e. 4 suits.
Total no. of ways of drawing 4 honour cards of same suit = 4
∴ Required probability = \(\frac{4}{{ }^{52} \mathrm{C}_4}\) = \(\frac{4}{\frac{52 \times 51 \times 50 \times 49}{24}}\) = \(\frac{4}{270725}\)

Question 13.
In a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously. Find the probability that
(i) both the tickets drawn have prime numbers ;
(ii) none of the tickets drawn has prime number ;
(iii) a ticket has prime number.
Solution:
Total exhaustive cases = Total no. of ways of drawing 2 tickets out of 50 = ⁵⁰C₂
(i) Prime numbers from 1 to 50 are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
∴ Total no. of prime numbers from 1 to 50 = 15
Thus total favourable cases = Total no. of ways of drawing 2 tickets out of 15 = ¹⁵C₂
∴ required probability = \(\frac { favourable cases }{ Exhaustive cases }\) = \(\frac{{ }^{15} C_2}{{ }^{50} C_2}\) = \(\frac{15 \times 14}{50 \times 49}\) = \(\frac{3}{35}\)

(ii) Total numbers which are not prime from 1 to 50 = 50 – 15 = 35
Thus total favourable cases = Total no. ot ways of drawing 2 tickets when none of tickets drawn has prime number = ³⁵C₂
∴ required probability = \(\frac{{ }^{35} \mathrm{C}_2}{{ }^{50} \mathrm{C}_2}\) = \(\frac{35 \times 34}{50 \times 49}\) = \(\frac{17}{35}\)

(iii) Total no. of ways of drawing 2 tickets in which one ticket containing prime no. = ¹⁵C₁ × ³⁵C₁
∴ required probability = \(\frac{{ }^{15} C_1 \times{ }^{35} C_1}{{ }^{50} C_2}\) = \(\frac{15 \times 35 \times 2}{50 \times 49}\) = \(\frac{3}{7}\)

Question 14.
(i) Out of 9 outstanding students in a college, there are 4 boys and 5 girls. A team of 4 students is to be selected for a quiz programme. Find the probability that 2 are girls and 2 are boys.
(ii) Four people are chosen at random from a group consisting of 3 men, 2 women, and 3 children. Find the probability that out of four chosen people, exactly 2 are children ?
Solution:
(i) Given total no. of outstanding students = 9
∴ Total no. of ways of selecting 4 students out of 9 = ⁹C₄
Total no. of ways of selecting 2 boys out of 4 be ⁴C₂ and no. of ways of selecting 2 girls out of 5 be ⁵C₂
∴ Total no. of ways of drawing 2 girls and 2 boys = ⁴C₂ × ⁵C₂

(ii) Total no. of peoples =3 + 2 + 3 = 8
∴ Total no. of ways of choosing 4 people out of 8 = ⁸C₄
Total no. of ways of choosing exactly 2 children = ³C₂ × ⁵C₂ = 3 × \(\frac { 5×4 }{ 2 }\) = 30
∴ required probability = \(\frac{30}{{ }^8 C_4}\) = \(\frac{30}{\frac{8 \times 7 \times 6 \times 5}{24}}\) = \(\frac { 3 }{ 7 }\)

Question 15.
A committee of 5 principals is to be selected from a group of 6 grant principals and 8 lady principals. If the selected is made randomly, find the probability that there are 3 lady principals and 2 gent principals.
Solution:
Total no. of principals = 6 + 8 = 14
∴ Total no. of ways of selecting 5 principals out of 14 = ¹⁴C₅
Total no. of ways of selecting 3 lady principals from 8 and 2 gents principals out of 6 = ⁸C₃ × ⁶C₂
∴ required probability = \(\frac{{ }^8 C_3 \times{ }^6 C_2}{{ }^{14} C_5}\) = \(\frac{\frac{8 \times 7 \times 6}{6} \times \frac{6 \times 5}{2}}{\frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2}}\) = \(\frac { 60 }{ 143 }\)

Question 16.
A bag contains tickets numbered 1 to 20. Two tickets are drawn. Find the probability that both numbers are odd.
Solution:
Total no. of ways of drawing 2 tickets out of 20 = ²⁰C₂
since there are 10 odd numbers from 1 to 20.
∴ Total no. of ways of drawing 2 tickets out of 10 = ¹⁰C₂
Thus required probability = \(\frac{{ }^{10} C_2}{{ }^{20} C_2}\) = \(\frac{10 \times 9}{20 \times 19}\) = \(\frac{9}{38}\)

Question 17.
A bag contains tickets numbered 1 to 30 . Three tickets are drawn at random from the bag. What is the probability that the maximum number of the selected tickets exceeds 25 ?
Solution:
Total no. of ways of drawing 3 tickets out of 30 = ³⁰C₃
Let A : event that none of the selected ticket bear number exceeding 25

Question 18.
(i) A room has 3 lamps. From a collection of 10 light bulbs of which 6 are no good, a person selects 3 at random and puts them in a socket. What is the probability, that he will have light ?
(ii) A has 3 shares in a lottery containing 3 prizes and 9 blanks ; B has two shares in lottery containing 2 prizes and 6 blanks. Compare their chances of success.
Solution:
(i) Total no. of light bulbs = 10
no. of no good bulbs = 6
Thus, no. of good bulbs = 10 – 6 = 4
Total exhaustive cases = total no. of ways of selecting 3 out of 10 = ¹⁰C₃
∴ required probability = 1 – P (no good bulb)

Question 19.
(i) There are n letters and n addressed envelopes. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelope ?
(ii) Three letters are written to different persons, and the addresses on the three envelopes are also written. Without looking at the addresses, find the probability that the letters go into the right envelopes.
Solution:
(i) Total no. of ways of planning n letters in n envelopes = n!
There is only one way in which all the letters placed correctly.
∴ probability of planning the letters in right envelopes = \(\frac{1}{n !}\)
Thus, required probability that all letters are not placed in right envelopes =1 – \(\frac{1}{n !}\)

(ii) The total number of ways of placing 3 letters in 3 envelopes = 3! = 6
There is only 1 way in which all the three letters placed correctly.
∴ required probability that the letters placed in right envelopes = \(\frac{1}{6}\)

Question 20.
The letters of word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ?
Solution:
The total no. of ways in which all the letters of the word SOCIETY are placed in a row = 7!
Total no. of vowels in word society =3 {O, I, E}
When all three vowels come together they form a group. ∴ 1 group and 4 consonants can be arranged in 5! ways and 3 vowels arranged themselves in 3! ways.
Thus total no. of ways in which three vowels come together = \(\lfloor 5\) × \(\lfloor 3\)
∴ required probability = \(\frac{3 ! \times 5 !}{7 !}\) = \(\frac{6}{7 \times 6}\) = \(\frac{1}{7}\)

Question 21.
In a random arrangement of the letters of the word “COMMERCE”, find the probability that all vowels come together.
Solution:
In word COMMERCE, there are 2E’s, 2M’s, 2C’s, 1O and 1R
∴ Total no. of ways in which letters of word COMMERCE arranged themselves = \(\frac{8 !}{2 ! 2 ! 2 !}\)
since there are 3 vowels (O, E, E) and came together to form a group. Thus, this group and 5 consonants arranged themselves in \(\frac{6 !}{2 ! 2 !}\) and 3 vowels can be arranged themselves in \(\frac{3 !}{2 !}\) .
Thus total no. of favourable cases = \(\frac{6 !}{2 ! 2 !} \times \frac{3 !}{2 !}\)
∴ required probability = \(\frac{\frac{6 !}{2 ! 2 !} \times \frac{3 !}{2 !}}{\frac{8 !}{2 ! 2 ! 2 !}}\) = \(\frac{6 ! \times 3 !}{8 !}\) = \(\frac{3}{28}\)

Question 22.
Given a group of 4 persons, find the probability that
(i) no two of them have their birthdays on the same day,
(ii) ail of them have same birthday. [Ignore the existence of a leap year]
Solution:
(i) The first person can have birthday on any of 365 days.
The second person can also have birthday on any of 365 days and so on.
Thus the total no. of ways of choosing the birthday for the four persons = (365)⁴
(i) Since no two persons have their birthday on the same day. Thus for the first person we have 365 choices and second person have 364, third person have 363 and 4th person have 362 choices.
Thus, the total no. of ways in which all the four persons have their birthdays on different days = 365 × 364 × 363 × 362
∴ required probability = \(\frac{365 \times 364 \times 363 \times 362}{(365)^4}\) = \(\frac{{ }^{364} \mathrm{P}_3}{(365)^2}\)

(ii) Since all the four persons have birthday on same day and that day can be any day out of365 days
∴ Total no. of such ways = 365
∴ Required probability = \(\frac{365}{(365)^4}\) = \(\frac{1}{(365)^3}\)

Utilizing S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(c) as a study aid can enhance exam preparation.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(c)

Question 1.
A die is thrown once. Find
(i) P (an ace),
(ii) P (an even number),
(iii) P (a number < 3),
(iv) P (a number ≥ 4),
(v) P (a number < 7),
(vi) P (a number > 8).
Solution:
When a die is drawn once.
Then sample space S = {1, 2, 3, 4, 5, 6} ∴ n (S) = 6
(i) P (an ace) = \(\frac { 1 }{ 6 }\)

(ii) P (an even number) = \(\frac { 3 }{ 6 }\)
Here favourable cases are (2, 4, 6}

(iii) E : getting a no. < 3 = {1, 2} ∴ n (E) = 2
∴ P (a number < 3) = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

(iv) F : getting a number ≥ 4 = {4, 5, 6} ∴ n(F) = 3
∴ P (getting a number ≥ 4) = \(\frac{n(\mathrm{F})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

(v) G : getting a number < 7 = {1, 2, 3, 4, 5, 6} ∴ n (G) = 6
∴ required probability = \(\frac{n(\mathrm{G})}{n(\mathrm{~S})}\) = \(\frac { 6 }{ 6 }\) = 1

(vi) H : getting a number > 8, since there is no number which is greater than 8 in case of dice ∴ n (H) = 0
Thus required probability = \(\frac{n(\mathrm{H})}{n(\mathrm{~S})}\) = \(\frac { 0 }{ 6 }\) = 0

Question 2.
A card is drawn from a well shuffled pack of 52 cards. Find the probability of (i) an ace (ii) a spade (iii) a black card (iv) a face card (v) Jack, queen or king (vi) 3 of heart or diamond.
Solution:
Total no. of outcomes = 52 = n(S)
(i) A : getting on ace card n (A) = 4
since there are 4 are cards.
Thus required probability = \(\frac{n(\mathrm{A})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)

(ii) No. of spade cards = 13 = no. of favourable outcomes
∴ required probability = \(\frac { no. of favourable outcomes }{ Total no. of outcomes }\) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)

(iii) No. of black cards = 13 + 13 = 26 = no. of favourable outcomes [since there 13 spade and 13 club cards]
∴ required probability = \(\frac { no. of favourable outcomes }{ Total no. of outcomes }\) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

(iv) No. of favourable outcomes = No. of face cards = 12
∴ required probability = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)

(v) No. of favourable outcomes = No. of king, queen or Jack cards = 4 + 4 + 4= 12
∴ required probability = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)

(vi) No. of favourable outcomes = No. of 3 of heart or diamond = 2
∴ required probability = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)

Question 3.
One card is drawn from a pack of 52 cards being equally likely to be drawn. Find the probability of (i) the card drawn to be red (ii) the card drawn to be a king (iii) the card drawn to be red and a king (iv) the card drawn to be either red or a king.
Solution:
Total no. of outcomes = total no. of cards = 52 = n (S)
(i) Let A : card drawn to be red
since there are 26 red cards ∴ n (A) = 26
Thus required probability = \(\frac{n(\mathrm{A})}{n(\mathrm{~S})}\) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

(ii) Let B : card drawn to be king ∴ n (B) = 4
Thus required probability = \(\frac{n(\mathrm{B})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)

(iii) Let C : card drawn to be red and king
since there are 2 red king cards, one of heart and other of diamond ∴ n (C) = 2
Thus required probability = \(\frac{n(\mathrm{C})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)

(iv) Let E : drawn card be either red or king
since there are 26 red cards includes 2 red kings and also we have 2 black king cards ∴ n (E) = 26 + 2 = 28
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 28 }{ 52 }\) = \(\frac { 7 }{ 13 }\)

Question 4.
A book contains 100 pages. A page is chosen at random. What is the chance that the sum of digits on the page is equal to 9 ?
Solution:
Total number of outcomes = 100 = n (S)
A : sum of the digits on the page is equal to 9 = {9, 18, 27, 36, 45, 54, 63, 72, 81, 90}
∴ n(A) = 10
Thus required probability = \(\frac{n(\mathrm{A})}{n(\mathrm{~S})}\) = \(\frac { 10 }{ 100 }\) = \(\frac { 1 }{ 10 }\)

Question 5.
From 25 tickets, marked with the first 25 numerals, one is drawn at random. Find the probability that (i) it is a multiple of 5 or 7 (ii) it is a multiple of 3 or 7.
Solution:
Total no. of outcomes = n (S) = 25
(i) Let E : drawn ticket is a multiple of 5 or 7 = {5, 7, 10, 14, 15, 20, 21, 25} ∴ n (E) = 8
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 8 }{ 25 }\)

(ii) Let F : drawn ticket is a multiple of 3 or 7 = {3, 6, 7, 9, 12, 14, 15, 18, 21, 24} ∴ n (F) = 10
Thus required probability = \(\frac{n(\mathrm{F})}{n(\mathrm{~S})}\) = \(\frac { 10 }{ 25 }\) = \(\frac { 2}{ 5 }\)

Question 6.
What is the probability that a number selected from the numbers 1,2, 3,…, 25 is a prime number ? You may assume that each of the 25 numbers is equally likely to be selected.
Solution:
Given total no. of outcomes = n (S) = 25
Let E : selected no. is a prime number = {2, 3, 5, 7, 11, 13, 17, 19, 23}
∴ n(E) = 9
Thus required probability of getting a prime no. = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 9 }{ 25 }\)

Question 7.
In a simultaneous throw of two coins find the probability of (i) two heads, (ii) exactly one tail, (iii) at least one tail.
Solution:
When two coins thrown simultaneously
Then S = {HH, HT, TH, TT} ∴ n (S) = 4
(i) E : getting two heads = {HH} ∴ n (E) = 1
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 4 }\)

(ii) F : getting exactly one tail = {HT, TH} ∴ n (F) = 2
Thus required probability = \(\frac{n(\mathrm{F})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 4 }\) = \(\frac { 1 }{ 2 }\)

(iii) G : getting atleast one tail = {HT, TH, TT} ∴ n (G) = 3
Thus required probability = \(\frac{n(\mathrm{G})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 4 }\)

Question 8.
(i) In a single throw of two dice, find the probability of getting a total of 10 or 11.
(ii) Two dice are thrown. Find the probability of getting an odd number on one die and a multiple of 3 on the other.
(iii) Find the probability of getting a sum as 6 when two dice are thrown simultaneously.
(iv) Two dice are thrown simultaneously. Find the probability of getting a multiple of 3 as the sum.
(v) Find the probability of getting the sum as a prime number when two dice are thrown together.
(vi) In a single throw of two dice, find the probability of throwing
(a) a number > 4 on each die ;
(b) an odd number on one die and 5 on the other ;
(c) a multiple of 2 on one and a multiple of 3 on the other.
(d) an even number as the sum ;
(e) an odd number as the sum.
(vii) In a single throw of two dice, what is the probability of
(a) two aces
(b) at least one ace
(c) doublets
(d) a total less than 10
(e) a total of 11
(f) a total of 12
(g) a total of at least 10
(h) a doublet of even number
(viii) Find the probability of getting the product a perfect square (square of a natural number), when two dice are thrown together.
Solution:
(i) In a single throw of two dice
Total no. of outcomes = 6 × 6 = 36 = n(S)
Let E : getting a total of 10 or 11 = {(5, 5), (4, 6), (6, 4), (5, 6), (6, 5)} ∴ n (E) = 5
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac{5}{36}\)

(ii) Here total no. of outcomes = 6² = 36 = n (S)
Let F : getting an odd number on one dice and a multiple of 3 on the other
= {(1, 3), (1, 6), (3, 3), (3, 6), (5, 3), (5, 6), (3, 1), (6, 1), (6, 3), (3, 5), (6, 5)}
∴ n (F) = 11
Thus required probability = \(\frac{n(\mathrm{F})}{n(\mathrm{~S})}\) = \(\frac{11}{36}\)

(iii) Here total no. of outcomes = n (S) = 36
G : getting a sum as 6 = {(1, 5), (2,4), (3, 3), (4, 2), (5, 1)} ∴ n (G) = 5
Thus required probability = \(\frac{n(\mathrm{G})}{n(\mathrm{~S})}\) = \(\frac{5}{36}\)

(iv) Total no. of outcomes = 6² = 36 = n (S)
Let A : getting a multiple of 3 as the sum
= {(1,2), (1, 5), (2, 1), (2, 4), (3,3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)} ∴ n (A) = 12
Thus required probability = \(\frac{n(\mathrm{A})}{n(\mathrm{~S})}\) = \(\frac{12}{36}\) = \(\frac{1}{3}\)

(v) When two dice are thrown together then n (S) = 6 × 6 = 36
Let E : getting the sum as a prime number
= {(1, 1), (1,2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
Therefore n (E) = 15
∴ required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac{15}{36}\) = \(\frac{5}{12}\)

(vi) In a single throw of two dice then total no. of outcomes n (S) = 6² = 36
(a) Let A : getting a number > 4 on each dice = {(5, 5), (5, 6), (6, 5), (6, 6)} ,∴ n (A) = 4
Thus required probability = \(\frac{n(\mathrm{A})}{n(\mathrm{~S})}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\)

(e) Let A : getting a total of 11 = {(5, 6), (6, 5)} ∴ n (A) = 2
Thus required probability = \(\frac{n(\mathrm{A})}{n(\mathrm{~S})}\) = \(\frac{2}{36}\) = \(\frac{1}{18}\)

(f) Let B : getting a total of 12 = {(6, 6)} ∴ n (B) = 1
∴ Thus required probability = \(\frac{n(\mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{1}{36}\)

(g) Let C : getting a total of atleast 10 = {(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)} ∴ n (C) = 6
∴ required probability = \(\frac{n(\mathrm{C})}{n(\mathrm{~S})}\) = \(\frac{6}{36}\) = \(\frac{1}{6}\)

(h) Let D : getting a doublet of even number = {(2, 2), (4, 4), (6, 6)} ∴ n (D) = 3
∴ required probability = \(\frac{n(\mathrm{D})}{n(\mathrm{~S})}\) = \(\frac{3}{36}\) = \(\frac{1}{12}\)

(vii) When two dice are thrown together
Then total no. of outcomes = n (S) = 6² = 36
Let E : getting the product as perfect square
= {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)} ∴ n (E) = 8
thus, required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac{8}{36}\) = \(\frac{2}{9}\)

Question 9.
In a single throw of three dice, find the probability of getting a total of 17 or 18.
Solution:
In a single throw of three dice
total no. of outcomes = n (S) = 6³ = 216
Let E : getting a total of 17 or 18 = {(5, 6, 6), (6, 5, 6), (6, 6, 5), (6, 6, 6)} ∴ n (E) = 4
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac{4}{216}\) = \(\frac{1}{54}\)

Question 10.
In a single throw of three dice, find the probability of getting
(i) a total of 5
(ii) a total of at most 5
(iii) a total of atleast 5
(iv) the same number on all the dice
(v) not getting the same number on all the dice.
Solution:
In a single toss of three dice
Then total no. of outcomes = n (S) = 6³ = 216
(i) Let A: getting a total ofS = {(l, 1,3), (1,3, 1), (3, 1, 1), (1,2,2), (2, 1,2), (2, 2, 1)}
∴ n (A) 6
Thus required probability = \(\frac{n(\mathrm{A})}{n(\mathrm{~S})}\) = \(\frac{6}{216}\) = \(\frac{1}{36}\)

(ii) Let B : getting a total of atmost 5 = {(1, 1, 1), (1, 2, 1), (2, 1, 1), (1, 1, 2), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1,2), (1,2, 2)}
∴ n (B) = 10
Thus required probability = \(\frac{n(\mathrm{B})}{n(\mathrm{~S})}\) = \(\frac{10}{216}\) = \(\frac{5}{108}\)

(iii) Let C : getting atmost 4 = {(1, 1, 1), (1, 2, 1), (2, 1, 1), (1, 1, 2)} ,∴ n (C) = 4
Thus probability of getting atmost 4 = \(\frac{n(\mathrm{C})}{n(\mathrm{~S})}\) = \(\frac{4}{216}\)
∴ required probability of getting a total of atleast 5 = 1 – P (getting a total of atmost 4)
= 1 – \(\frac{4}{216}\) = \(\frac{212}{216}\) = \(\frac{53}{54}\)

(iv) Let D : getting same no. on all the dice = {(1, 1, 1), (2, 2,2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6,6, 6)} ∴ n (D) = 6
Thus required probability = \(\frac{n(\mathrm{D})}{n(\mathrm{~S})}\) = \(\frac{6}{216}\) = \(\frac{1}{36}\)

(v) required probability = 1 – P (getting same number on all dice) = 1 – \(\frac{1}{36}\) = \(\frac{35}{36}\)

Question 11.
There are four events E₁, E₂, E₃, and E₄ one of which must and only one can happen. The odds are 2 : 5 in favour of E₁, 3 : 4 in favour of E₂ and 1 : 3 in favour of E₃. Find the odds against E₄.
Solution:
Given odds in favour of E₁ be 2 : 5 ∴ P(E₁) = \(\frac{2}{2+5}\) = \(\frac{2}{7}\)
Also, odds in favour of E₂ be 3 : 4. ∴ P(E₂) = \(\frac{3}{3+4}\) = \(\frac{3}{7}\)
Odds in favour of E₃ be 1 : 3. ∴ P(E₃) = \(\frac{1}{1+3}\) = \(\frac{1}{4}\)
∴ P (E₄) = 1 – P (E₁) – P (E₂) – P (E₃) = 1 – \(\frac{2}{7}\) – \(\frac{3}{7}\) – \(\frac{1}{4}\) = \(\frac{28-8-12-7}{28}\) = \(\frac{1}{28}\)
Thus \(\mathrm{P}\left(\overline{\mathrm{E}}_4\right)\) = 1 – P(E₄) = 1 – \(\frac{1}{28}\) – \(\frac{27}{28}\)
∴ odds against E₄ = \(\mathrm{P}\left(\overline{\mathrm{E}}_4\right)\) : P (E₄ ) = \(\frac{27}{28}\) : \(\frac{1}{28}\) = 27 : 1

Question 12.
In a simultaneous toss of 4 coins, what is the probability of getting exactly 3 heads ?
Solution:
In a simultaneous throw of 4 coins
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
∴ n (S) = 2⁴ = 16
Let E : getting exactly 3 heads = {HHHT, HHTH, HTHH, THHH}, ∴ n (E) = 4
Thus required probability = \(\frac{n(\mathrm{D})}{n(\mathrm{~S})}\) = \(\frac{4}{16}\) = \(\frac{1}{4}\)

Interactive S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(b) engage students in active learning and exploration.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(b)

Question 1.
What is the probability of getting :

Solution:
(i) Total no. of outcomes = 2
no. of favourable outcomes = 1 {H}
∴ required probability of getting head = \(\frac { 1 }{ 2 }\)

(ii) When a die is known then S = {1, 2, 3, 4, 5, 6}
Then total no. of outcomes = n (S) = 6
E : getting a 6 ∴ n (E) = 1
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 6 }\)

(iii) Here S = { 2, 3, 4, 5, 6, 7}
Then n (S) = 6
E : getting a prime no. = {2, 3, 5, 7} ∴ n (E) = 4
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)

Question 2.
Ramesh chooses a date at random in April for a party. Calculate the probability that he chooses :
(i) a Saturday
(ii) a Sunday
(iii) a Saturday or a Sunday.

Solution:
(i) Here n (S) = 30
Let E : event that he chooses a Saturday ∴ n (E) = 4
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 30 }\) = \(\frac { 2 }{ 15 }\)

(ii) F : a Sunday is chosen = {4, 11, 18, 25}
∴ (F) = 4
Thus required probability = \(\frac{n(\mathrm{F})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 30 }\)

(iii) G : a Saturday or a Sunday is chosen = {3, 10, 17, 24, 4, 11, 18, 25}
∴ n (G) = 8
Thus required probability = \(\frac{n(\mathrm{G})}{n(\mathrm{~S})}\) = \(\frac { 8 }{ 30 }\) = \(\frac { 4 }{ 15 }\)

Question 3.
A normal die is rolled. Calculate the probability that the number on the uppermost face when it stops rolling will be

(i) 5
(ii) not 5
(iii) an odd number
(iv) a prime number
(v) a 3 or a 4
(vi) a 1 or a 2 or a 3 or a 4.
Solution:
When a die is rolled then S = {1, 2, 3, 4, 5, 6} ∴ n (S) = 6
(i) A : getting a 5 ∴ n (A) = 1
∴ required probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 6 }\)

(ii) B : getting not 5 = { 1, 2, 3, 4, 6} ∴ n (B) = 5
This required probability = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 5 }{ 6 }\)

(iii) C : getting an odd no. = {1, 3, 5} ∴ n (C) = 3
Thus, required probability = \(\frac{n(\mathrm{~C})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

Question 4.
Nine playing cards are numbered 2 to 10. A card is selected from them at random. Calculate the probability that the card will be :
(i) an odd number
(ii) a multiple of 4.
Solution:
Here S = {2, 3, 4, 5, 6, 7, 8, 9, 10} ∴ n (S) = 9
(i) E : getting an odd numbered card = {3, 5, 7, 9} ∴ n (E) = 4
Thus required probability = \(\frac{n(\mathrm{~E})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 9 }\)

(ii) F : getting a card with multiple of 4 = {4, 8} ∴ n (F) = 2
Thus required probability = \(\frac{n(\mathrm{~F})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 9 }\)

Question 5.
Nine counters numbered 2 to 10 are put in a bag. One counter is selected at random. What is the probability of getting a counter with :
(i) a number 5
(ii) an odd number
(iii) not an odd number
(iv) a prime number
(v) a square number
(vi) a multiple of 3 ?
Solution:
Here n (S) = total no. of outcomes = 9
(i) E : getting a counter with no. 5 ∴ n (E) = 1
Thus required probability = \(\frac{n(\mathrm{~E})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 9 }\)

(ii) F : getting a counter with an odd number = {3, 5, 7, 9} ∴ n (F) = 4
Thus required probability = \(\frac{n(\mathrm{~F})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 9 }\)

(iii) G : getting a counter with not an odd number = {2, 4, 6, 8, 10} ∴ n (G) = 5
Thus required probability = \(\frac{n(\mathrm{~F})}{n(\mathrm{~S})}\) = \(\frac { 5 }{ 9 }\)

(iv) H : getting a prime number = {2, 3, 5, 7} ∴ n (H) = 4
Thus required probability = \(\frac{n(\mathrm{~H})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 9 }\)

(v) A : getting a counter with a square number = {4, 9} ∴ n(A) = 2
Thus required probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 9 }\)

(vi) B : getting a counter with a multiple of 3 = {3, 6, 9} ∴ n (B) = 3
Thus required probability = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)

Question 6.
A die is rolled. If the outcome is an even number, what is the probability that it is a prime number.
Solution:
It is given that outcome is an even number on rolling a die then S = {2, 4, 6} ∴ n (S) = 3
A : getting a prime no. = {2} ∴ n (A) = 1
Thus required probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 3 }\)

Question 7.
A bag contains 20 coloured balls. 8 are red, 6 are blue, 3 are green, 2 are white and 1 is brown. A ball is chosen at random from the bag. What is the probability that the ball chosen is :
(i) blue
(ii) not blue
(iii) brown
(iv) not brown
(v) blue or red
(vi) red or green
(vii) green or white or brown ?
Solution:
Given no. of red balls = 8 ;
no. of green balls = 3 ;
no. of blue balls = 6 ;
no. of white balls = 2 ;
no. of brown balls = 1
∴ Total no. of balls = 20 = n (S)
(i) required probability of getting a blue ball = \(\frac { no. of favourable outcomes }{ n(S) }\) = \(\frac { 6 }{ 20 }\) = \(\frac { 3 }{ 10 }\)

(ii) required probability of getting not a blue ball = \(\frac{(20-6)}{20}\) = \(\frac { 14 }{ 20 }\) = \(\frac { 7 }{ 10 }\)

(iii) Required probability of getting a brown ball = \(\frac { 1 }{ 20 }\)

(iv) Required probability of getting not a brown ball = \(\frac { 20-1 }{ 20 }\) = \(\frac { 19 }{ 20 }\)

(v) Required probability of getting blue or red ball = \(\frac {8+6}{ 20 }\) = \(\frac { 14 }{ 20 }\) = \(\frac { 7 }{ 10 }\)

(vi) Required probability of getting red or green ball = \(\frac {8+3}{ 20 }\) = \(\frac { 11 }{ 20 }\)

(vii) Required probability of getting green or white or brown ball = \(\frac {3+2+1}{ 20 }\) = \(\frac { 6 }{ 20 }\) = \(\frac { 3 }{ 10 }\)

Question 8.
A bag contains 20 balls. These are of three different colours : green, red and blue. A ball is chosen at random from the bag. The probability of a green ball is \(\frac { 1 }{ 4 }\) The probability of a red ball is \(\frac { 2 }{ 5 }\).
(i) What is the probability of a blue ball ?
(ii) How many balls are red ?
(iii) How many balls are green ?
(iv) How many balls are blue ?
Solution:
Given total no. of balls = 20 = n (S)
given probability of getting green ball = P (G) = \(\frac { 1 }{ 4 }\)
probability of getting green ball = P (R) = \(\frac { 2 }{ 5 }\)

(i) Required probability of getting a blue ball = 1 – P (G) – P (R)
= 1 – \(\frac { 1 }{ 4 }\) – \(\frac { 2 }{ 5 }\) = \(\frac { 20-5-8 }{ 20 }\) = \(\frac { 7 }{ 20 }\)

(ii) Let x be the no. of red balls in a bag
Thus probability of getting a red ball = \(\frac { x }{ 20 }\) ∴ \(\frac { x }{ 20 }\) = \(\frac { 2 }{ 5 }\) ⇒ x = 8

(iii) Let y be the no. of green balls in a bag.
Then probability of getting a green ball = \(\frac { y }{ 20 }\) ∴ \(\frac { y }{ 20 }\) = \(\frac { 1 }{ 4 }\) ⇒ y = 5
Thus required no. of green balls in a bag = 5

(iv) ∴ required no. of blue balls = 20 – 8 – 5 = 7

Question 9.
A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Solution:
When a pair of dice is thrown
Then S = {(1, 1), (1,2), (1,3), (1,4), (1, 5), (1,6), (2, 1), (2,2), (2, 3), (2,4), (2, 5), (2,6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5,2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} ∴ n (S) = 36
Let A : getting a sum of 10 or more if 5 appears on the first die = {(5, 5), (5, 6)} ∴ n (A) = 2
required probability = \(\frac { 2 }{ 36 }\) = \(\frac { 1 }{ 18 }\)

Question 10.
A match can be won, drawn or lost. One week a school is to play two matches. Draw a tree diagram to show all the possible outcomes and list them. If the outcomes are equally likely, calculate the probability that:
(i) both matches are won ;
(ii) one match is drawn ;
(iii) at least one match is drawn ;
(iv) no match is lost;
(v) both matches are not lost.
Solution:
Since a match can be won, drawn or lost

S = {WW, WD, WL, DW, DD, DL, LW, LD, LL}
Thus n(S) = 9

(i) Let E : both matches are won = {WW} ∴ n (E) = 1
Thus required probability = \(\frac{n(\mathrm{~E})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 9 }\)

(if) F : one match is drawn = {WD, DW, DL, LD} ∴ n (F) = 4
Thus required probability = \(\frac{n(\mathrm{~F})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 9 }\)

(iii) G : atleast one match is drawn = {WD, DW, LD, DL, DD} ∴ n (G) = 5
Thus required probability = \(\frac{n(\mathrm{~G})}{n(\mathrm{~S})}\) = \(\frac { 5 }{ 9 }\)

(iv) A : No match is lost = {WW, WD, DW, DD} ∴ n (A) = 4
Thus, required probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 4 }{ 9 }\)

(v) B : both matches are not lost = {WW, WD, DW, DD, DL, LD, WL, LW} ∴ n (B) = 8
Thus, required probability = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 8 }{ 9 }\)

Question 11.
The ace, king, queen, jack and ten from both the spades and hearts suits are placed in two separate piles and one card is taken from each pile : Draw the sample space diagram and find the probability that:

(i) both cards will be kings ;
(ii) both of the cards could be either ace or a king ;
(iii) both cards will be a pair ;
(iv) at least one card will be an ace ;
(v) neither card will be a 10 ;
(vi) neither card will be a king or jack;
(vii) one card will be a spade ;
(viii) both cards will be hearts.
Note. Description of normal pack of cards (52)

Each of the colours, hearts, diamonds, clubs, spades, is called a suit. Kings, queens and jacks are called face cards. There are 12 face cards in a normal pack of cards.
Solution:
Space diagram is given as under :

∴ Sample space S = {AA, AK, AQ, AJ, A10, KA, KK, KQ, KJ, K10, QA, QK, QQ, QJ, Q10, JA, JK, JQ, JJ, J10, 10A, 10K, 10Q, 10J, 1010} ∴ n (S) = 5 × 5 = 25

(i) Let A : both drawn cards are kings = {KK} ∴ n (A) = 1
Thus required Probability = \(\frac{n(\mathrm{~A})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 25 }\)

(ii) Let B : both cards could be either ace or a king = {AA, KK} ∴ n (B) = 2
Thus required Probability = \(\frac{n(\mathrm{~B})}{n(\mathrm{~S})}\) = \(\frac { 2 }{ 25 }\)

(iii) Let C : both drawn cards will be a pair = {AA, KK, QQ, JJ, 1010} ∴ n (C) = 5
Thus required Probability = \(\frac{n(\mathrm{~C})}{n(\mathrm{~S})}\) = \(\frac { 5 }{ 25 }\) = \(\frac { 1 }{ 5 }\)

(iv) E : atleast one card will be an ace = {AK, AA, AQ, AI, A10, KA, QA, JA, 10A} ,∴ n(E) = 9
Thus required Probability = \(\frac{n(\mathrm{~E})}{n(\mathrm{~S})}\) = \(\frac { 9 }{ 25 }\)

(v) F : neither card will be 10 = {AK, AA, AJ, AQ, KA, KK, KQ, KJ, QA, QK, QJ, QQ, JA, JK, JQ, JJ}
Thus n (F) = 16
∴ required probability = \(\frac{n(\mathrm{~F})}{n(\mathrm{~S})}\) = \(\frac { 16 }{ 25 }\)

(vi) G : neither card will be King or Jack = {AA, AQ, A10, QA, QQ, Q10, 10A, 10Q, 1010}
∴ n (G) = 9
Thus required probability = \(\frac{n(\mathrm{~G})}{n(\mathrm{~S})}\) = \(\frac { 9 }{ 25 }\)

(vii) H : one card will be spade ∴ n (H) = 25
∴ required probability = \(\frac{n(\mathrm{~H})}{n(\mathrm{~S})}\) = \(\frac { 25 }{ 25 }\) = 1

(viii) I : both cards will be heart cards
out of two drawn cards, one always must be from spade and one from heart so both cards can never be from heart suit.
∴ n(I) = 0
Thus required probability = \(\frac{n(\mathrm{~I})}{n(\mathrm{~S})}\) = \(\frac { 0 }{ 25 }\) = 0

Interactive S Chand ISC Maths Class 11 Solutions Chapter 22 Probability Ex 22(a) engage students in active learning and exploration.

S Chand Class 11 ICSE Maths Solutions Chapter 22 Probability Ex 22(a)

Question 1.
What do you mean by Random Experiment ? Give two illustrations. Define sample space associated with a random experiment. Give an example.
Solution:
Random Experiment: An experiment whose all outcomes are known in advance but outcomes of experiment cannot be predictable.
e.g. : Tossing a coin, its outcomes are known i.e. either head or tail but we can’t predict the outcome i.e. on tossing a coin, we can’t predict whether head comes or tail. e.g. : throwing a dice, outcomes are known i.e. {1, 2, 3, 4, 5, 6} but we can’t predict the outcomes which number comes.
Sample space : The set of all outcomes of a random experiment is called sample space.
e.g. : On tossing a coin, outcomes are either head or tail.
∴ Sample space = {H, T}

Question 2.
What is the resulting sample space if
(i) one coin is tossed ;
(ii) two coins are tossed simultaneously ;
(iii) three coins are tossed simultaneously ?
Solution:
(i) When one coin is tossed
Then sample space S = {H, T}
(ii) Two coins are tossed simultaneously
Then S = {HH, HT, TH, TT}

Question 3.
Describe the sample space of this experiment:
(i) One die is rolled ;
(ii) Two dice are rolled.
Solution:
(i) When one die is rolled
Then sample space S = {1, 2, 3,4, 5, 6}
(ii) When two dice are rolled
Then S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Question 4.
Describe the sample space :
(i) A coin is tossed twice. If the second throw results in a tail, a die is thrown.
(ii) A coin is tossed twice. If the second throw results in a head, a die is thrown, otherwise a coin is tossed.
(iii) A coin is tossed. If it results in a head, a die is thrown. If the die shows up an even number, the die is thrown again.
Solution:
(i) S = {HH, HT1, HT2, HT3, HT4, HT5, HT6, TT1, TT2, TT3, TT4, TT5, TT6}
(ii) S = {HTH, TTH, TTT, HTT, HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6}
(iii) S = {T, HI, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}

Question 5.
A five-sided spinner is spun and a coin is tossed.
(i) Show the combined outcomes in a space diagram and in a tree diagram.
(ii) List the combined outcomes and state the number of equally likely combined outcomes.

Solution:

(ii) {(H , 1), (H, 2), (H, 3), (H, 4), (H, 5), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5)}
∴ required total no. of outcomes = 10

Question 6.
In a bag there are three balls ; one red, one blue and one yellow. A ball is selected, the colour is recorded and the ball is replaced. A second ball is then selected and the colour is recorded.
(i) Show in a space diagram and in a tree diagram all the possible combined outcomes.
(ii) List these combined outcomes and state the number of equally likely combined outcomes.
Solution:

(ii) {(R. R), (R, B), (R, Y), (B, R), (B, B), (B, Y), (Y, R), (Y, B), (Y, Y)}

Question 7.
Satish and Mukesh who live in London wish to go on a holiday to France. They can travel to the coast by car, coach or train, and then cross the channel by ferry, train, helicopter or hovercraft.
(i) In a space diagram and in a tree diagram show all the combined outcomes of the different ways they could travel to France.
(ii) How many different ways could they travel ?
Solution:

(ii) Thus required no. of ways = 4 + 4 + 4 = 12

Question 8.
From a group of 2 men and 3 women, two persons are selected. Describe the sample space of the experiment. If E is the event in which one man and one woman are selected, then which are the cases favourable to E ?
Solution:
Given we have group of 2 men {M₁, M₂} and 3 women {W₁, W₂, W₃}
∴ Sample space = {M₁M₂, M₁W₁, M₁W₂, M₁W₃, M₂W₁, M₂W₂, M₂W₃, W₁W₂, W₂W₃, W₁W₃} Given E : event in which one man and one woman be selected
∴ favourable cases to E = {M₁W₁, M₁W₂, M₁W₃, M₂W₁, M₂W₂, M₂W₃}

Question 9.
A coin is tossed. If it results in a head, a coin is tossed, otherwise a die is thrown. Describe the following events:
(i) A: getting at least one head ;
(ii) B : getting an even number ;
(iii) C : Getting a tail;
(iv) D : getting a tail and an odd number.
Solution:
Sample space = {HH, HT, T1, T2, T3, T4, T5, T6}
(i) A = {HH, HT}
(ii) B = {T2, T4, T6}
(iii) C = {HT, Tl, T2, T3, T4, T5, T6}
(iv) D = {T1, T3, T5}

Question 10.
A coin and a die are tossed. Describe the following events.
(i) A : getting a head and an even number ;
(ii) B : getting a prime number ;
(iii) C : getting a tail and an odd number;
(iv) D : getting a head or a tail.
Solution:
When a coin and a dice are thrown
Then sample space S = {H1, H2, H3, H4, H5, H6, Tl, T2, T3, T4, T5, T6}
(i) A = {H2, H4, H6}
(ii) B = {H2, H3, H5, T2, T3, T5}
(iii) C = {T1, T3, T5}
(iv) D = {H1, H2, H3, H4, H5, H6, Tl, T2, T3, T4, T5, T6}

Question 11.
A fair coin is tossed. If it shows a head, we draw a ball from a bag consisting of 3 distinct red and 4 distinct black balls, if it shows a tail, we throw a fair die. Draw a tree diagram to show all the possible outcomes and obtain .he sample space. What are sets representing the following events:
(i) the ball drawn is black ;
(ii) the coin shows tail.
Solution:

∴ S = {HR₁, HR₂, HR₃, HB₁, HB₂, HB₃, HB₄, T₁, T₂, T₃, T₄, T₅, T₆)
(i) {(H, B₁), (H, B₂), (H, B₃), (H, B₄)}
(ii) {(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}

Question 12.
Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5. B is the event that at least one of the dice shows up a 3. Are the two events A and B (i) mutually exclusive, (ii) exhaustive ? Give arguments in support of your answer.
Solution:
When two dice are rolled
Then S = {(1, 1), (1,2), (1,3), (1,4), (1, 5), (1,6), (2,1), (2,2), (2,3), (2,4), (2, 5), (2,6), (3,1), (3,2), (3,3), (3,4), (3, 5), (3,6), (4, 1), (4,2), (4, 3), (4,4), (4, 5), (4, 6), (5, 1), (5,2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ A = {(1,4), (2, 3), (3, 2), (4,1)}
B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3)} .
Here A ∩ B = {(2, 3), (3, 2)} ≠ Φ
Thus A and B are not mutually exclusive events.
Here, A∪B ≠ S .
∴ A and B are not exhaustive events.

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