OP Malhotra Class 12 Maths Solutions Chapter 20 Theoretical Probability Distribution Ex 20(b)

Students appreciate clear and concise OP Malhotra ISC Class 12 Solutions Chapter 20 Theoretical Probability Distribution Ex 20(b) that guide them through exercises.

S Chand Class 12 ICSE Maths Solutions Chapter 20 Theoretical Probability Distribution Ex 20(b)

Question 1.
In a single throw of a die, if X denotes the number on its upper face, find the mean of X.
Answer:
Here X denotes the number on its upper face
P(X = 1) = \(\frac{1}{6}\) = P(X = 2)
= P(X = 3)
= P(X = 4) = P(X = 5)
= P(X = 6)
∴ Mean = µ
= \(\sum_{i=1}^6\) X_i P(X_i)
= 1 × \(\frac{1}{6}\) + 2 × \(\frac{1}{6}\) + 3 × \(\frac{1}{6}\) + 4 × \(\frac{1}{6}\) + 5 × \(\frac{1}{6}\) + 6 × \(\frac{1}{6}\)
= \(\frac{1}{6}\) (1 + 2 + 3 + 4 + 5 + 6)
= \(\frac{21}{6}\)
= \(\frac{7}{2}\)

Question 2.
A salesman wants to know the average number of units he sells per sales call. He chccks his past sales records and comes up with the following probabilities.

Sales (in units)	0	1	2	3	4	5
probability	0.15	0.20	0.10	0.05	0.30	0.20

What is the average number of units he sells per sale call ?
Answer:

xi	pi	pi xi
0 1 2 3 4 5	01.5 0.20 0.10 0.05 0.30 0.20	0 0.20 0.20 0.15 1.20 1.00
		∑ pi xi = 2.75

∴ Mean = ∑ p_i x_i = 2.75

Question 3.
Find µ and variance for the following probability distribution.
(i)

X	2	5
P(X)	0.4	0.6

(ii)

x	1	2	3	4
P(x)	0.4	0.3	0.2	0.1

(iii)

x	-2	-1	0	1	2
P(x)	0.2	0.3	0.3	0.1	0.1

(iv)

X	2	3	4
P(X)	p	p	1 – 2p

Answer:

X	P(X)	X2	X2 P(X)	X P(X)
2 5	0.4 0.6	4 25	1.6 1.5	0.8 3.0
			∑ X2 P(X)  = 16.6	∑ X P(X) = 3.8

∴ Mean = µ = ∑ X P(X) = 3.8
Variance σ² = ∑² = ∑ X² P(X) – µ²
= 16.6 – (3.8)² = 2.16

X	P(X)	X P(X)	X2 P(X)
1 2 3 4	0.4 0.3 0.2 0.1	0.4 0.6 0.6 0.4	0.4 1.2 1.8 1.6
		∑ X P(X) = 2.0	∑ X2 P(X) = 5

∴ Mean = µ = ∑ X P(X) = 2
Variance σ² = ∑² = ∑ X² P(X) – µ²
= 5² – (2)² = 5 – 4 = 1

X	P(X)	X P(X)	X P(X)
-2 -1 0 1 2	0.2 0.3 0.3 0.1 0.1	-0.4 -0.3 0 0.1 0.2	+0.8 +0.3 0 0.1 0.4
		∑ X P(X) = -0.4	∑ X2 P(X) = +1.6

∴ Mean = µ = ∑ X P(X) = -0.4
Variance σ² = ∑² = ∑ X² P(X) – µ²
= +1.6 – (-0.4)² = +1.6 – 0.16 = 1.44

X	P(X)	X P(X)	X P(X)
2 3 4	P P 1-2P	2P 3P 4(1-2P)	4P 9P 16(1- 2P)
		∑ X P(X) = 4 – 3p	∑ X2 P(X) = 16 – 19p

∴ Mean = µ = ∑ X P(X) = 4 – 3p
Variance σ² = ∑² = ∑ X² P(X) – µ²
= 16 – 19p – (4 – 3p)²
= 16 – 19p – 16 – 9p² + 24p
= 5p – 9p²

Question 4.
A die is tossed thrice. If getting ‘four’ is considered a success, find the mean and variance of probability distribution of the number of successes.
Answer:
When a die is thrown then
Sample space S = {1, 2, 3, 4, 5, 6}
p = probability of success = prob. of getting 4
= \(\frac{1}{6}\) ∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
Let X be the random variable denotes the number of successes then X can take values 0,1,2 and 3.
∴ P(X = 0) = P(no success ) = q q q
= \(\frac{5}{6}\) × \(\frac{5}{6}\) × \(\frac{5}{6}\) = \(\frac{125}{216}\)
P(X = 1) = P(1 success) = p q q + q p q + q q p
= \(\frac{1}{6}\) × \(\frac{5}{6}\) × \(\frac{5}{6}\) + \(\frac{5}{6}\) × \(\frac{1}{6}\) × \(\frac{5}{6}\) + \(\frac{5}{6}\) × \(\frac{5}{6}\) × \(\frac{1}{6}\)
= \(\frac{75}{216}\)
P(X = 2) = P(2 successes)
= p p q + p q p + q p p
= \(\frac{1}{6}\) × \(\frac{1}{6}\) × \(\frac{5}{6}\) + \(\frac{1}{6}\) × \(\frac{5}{6}\) × \(\frac{1}{6}\) + \(\frac{5}{6}\) × \(\frac{1}{6}\) × \(\frac{1}{6}\)
= \(\frac{15}{216}\)
P(X = 3) = P( all 3 successes) = p p p
= \(\frac{1}{6}\) × \(\frac{1}{6}\) × \(\frac{1}{6}\)
= \(\frac{1}{216}\)
The table of values for computation of µ and ∑² is given as under :

x	p	pi xi	pi xi2
0	\(\frac{125}{216}\)	0	0
1	\(\frac{75}{216}\)	\(\frac{75}{216}\)	\(\frac{75}{216}\)
2	\(\frac{15}{216}\)	\(\frac{30}{216}\)	\(\frac{60}{216}\)
3	\(\frac{1}{216}\)	\(\frac{3}{216}\)	\(\frac{9}{216}\)
		∑pi xi = \(\frac{108}{216}\)	∑pi xi2= \(\frac{144}{216}\)

∴ Mean = µ = ∑p_i x_i
= \(\frac{108}{216}\)
= \(\frac{1}{2}\)
and Variance = ∑²
= ∑p_i x_i² – µ²
= \(\frac{144}{216}\) – (\(\frac{1}{2}\))²
= \(\frac{6}{9}\) – \(\frac{1}{4}\)
= \(\frac{2}{3}\) – \(\frac{1}{4}\)
= \(\frac{8-3}{12}\)
= \(\frac{5}{12}\)

Question 5.
Is it possible for the random variable X to have the following distribution?

X	-2	-1	0	1
P(X)	0.2	0.3	0.4	0.1

If it is possible, find the mean and the variance of the random variable.
Answer:
Here P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) = 0.2 + 0.3 + 0.4 + 0.1 = 1.0
Thus the sum of probabilities of distribution of probabilities is 1 .
∴ given distribution is probability distribution.
The table of values for computation of µ and ∑² is given as under :

x	pi	pixi	pixi2
-2	0.2	-0.4	0.8
-1	0.3	-0.3	0.3
0	0.4	0	0
1	0.1	0.1	0.1

∴ Mean = µ = ∑p_i x_i = -0.6
and Variance σ²
= ∑p_i x_i² – µ²
= 1.2 – (-0.6)² = 1.2 – 0.36 = 0.84

Question 6.
Two cards are drawn successively with replacement from a well shuffled deck. Determine the probability distribution of number of kings. Find the mean and variance of the distribution.
Answer:
Let X denotes the number of kings drawn in two successive draws and X can take values 0,1 and 2 .
∴ P(X = 0) = probability of getting no king
= \(\frac{48}{52}\) × \(\frac{48}{52}\)
= \(\frac{144}{169}\)
P(X = 1) = P(one king)
= \(\frac{4}{52}\) × \(\frac{48}{52}\) + \(\frac{48}{52}\) × \(\frac{4}{52}\)
= \(\frac{24}{169}\)
P(X = 2) = P(Two kings in two successive draws)
= \(\frac{4}{52}\) × \(\frac{4}{52}\)
= \(\frac{1}{169}\)
The probability distribution of X is given below :

X	0	1	2
P(X)	\(\frac{144}{169}\)	\(\frac{24}{169}\)	\(\frac{1}{169}\)

∴ Mean = µ = ∑X P(X) = \(\frac{26}{169}\)
= \(\frac{2}{13}\)
and Variance = ∑² = ∑²; P(X) – µ²;
= \(\frac{28}{169}\) – (\(\frac{2}{13}\))²;
= \(\frac{28}{169}\) – \(\frac{4}{169}\)
= \(\frac{24}{169}\)

x	P(X)	Xp(X)	Xpx
0	\(\frac{144}{169}\)	0	0
1	\(\frac{24}{169}\)	\(\frac{24}{169}\)	\(\frac{24}{169}\)
2	\(\frac{1}{169}\)	\(\frac{2}{169}\)	\(\frac{4}{169}\)

Question 7.
Find the mean, the variance and standard deviation of the number of heads in a simultaneous toss of three coins.
Answer:
When three coins are thrown simultaneously
Then S = {H H H, HHT, HTH, HTT, THH, THT, TTH, TTT }
Let X be the random variable and denotes the no. of heads in three tosses of a coin.
Let p= prob. of success i . e. prob. of getting a head in single toss of coin = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Thus X can take values 0, 1, 2, 3
P(X = 0) = P(no head ) = q q q
= \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\)
P(X = 1) = P(1 head )
= P(HTT, THT, TTH ) = \(\frac{3}{8}\)
P(X = 2) = P(2 heads)
= P(HHT, HTH, THH)
= \(\frac{3}{8}\)
P(X = 3) = P(3 heads )
= P(HHH)
= \(\frac{1}{8}\)
The table of values for µ and ∑² is given as under :

x	P(X)	Xp(X)	Xpx
0	\(\frac{1}{8}\)	0	0
1	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)
2	\(\frac{3}{8}\)	\(\frac{6}{8}\)	\(\frac{12}{8}\)
3	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{9}{8}\)
		∑ X P(X) = \(\frac{12}{8}\)	∑ X2 P(X) = \(\frac{24}{8}\) = 3

∴ Mean = µ
= ∑ X P(X)
= \(\frac{12}{8}\)
= \(\frac{3}{2}\)
and
∑² = variance = ∑ X² P(X) – µ²
= 3 – (\(\frac{3}{2}\))²
= 3 – \(\frac{9}{4}\)
= \(\frac{3}{4}\)
= 0.75
and
S.D = ∑ = \(\sqrt{\frac{3}{4}}\)
= \(\frac{\sqrt{3}}{2}\)

Question 8.
Find the mean and standard deviation of the probability distribution of the numbers obtained when a card is drawn at random from a set of 7 cards numbered 1 to 7 .
Answer:
Let X denote the number on the card drawing from cards numbered 1 to 7
∴ X can take values 1, 2, 3, 4, 5, 6 and 7 .
∴ p = prob. of getting a card with number 1 = \(\frac{1}{7}\)
i.e. P(X = 1) = \(\frac{1}{7}\) = P(X = 2)
= P(X = 3)
= P(X = 4)
= P(X = 5) = P(X = 6)
= P(X = 7)
∴ µ = Mean = ∑ X P(X) = 1 × \(\frac{1}{7}\) + 2 × \(\frac{1}{7}\) + 3 × \(\frac{1}{7}\) + 4 × \(\frac{1}{7}\) +5 × \(\frac{1}{7}\) + 6 × \(\frac{1}{7}\) + 7 × \(\frac{1}{7}\)
= \(\frac{1}{7}\)[1 + 2 + 3 + 4 + 5 + 6 + 7]
= \(\frac{28}{7}\) = 4
and Variance = ∑² = ∑ X² P(X) – µ²
= 1² × \(\frac{1}{7}\) + 2² × \(\frac{1}{7}\) + 3² × \(\frac{1}{7}\) + 4² × \(\frac{1}{7}\) + 5² × \(\frac{1}{7}\) + 6² × \(\frac{1}{7}\) + 7² × \(\frac{1}{7}\) – 4²
= \(\frac{1}{7}\)[1 + 4 + 9 + 16 + 25 + 36 + 49] – 4
= \(\frac{1}{7}\) × 140 – 4 = 20 – 16 = 4
∴ S.D = ∑ = \(\sqrt{\text { Variance }}\)
= \(\sqrt{4}\)
= 2

Question 9.
A pair of dice is rolied twice. Let Z denote the number of times, a total of 9 is obtained. Find the mean and variance of the random variable X.
Answer:
When a pair of dice is rolled
Then total no. of outcomes = 6² = 36
p = prob. of success i.e. prob. of getting 9 with a single throw of pair of dice
= \(\frac{4}{36}\)
= \(\frac{1}{9}\) { [Here favourable cases are }{(3,6),(4,5),(5,4),(6,3)}]
∴q = 1 – p = 1 – \(\frac{1}{9}\) = \(\frac{8}{9}\)
Let X be the random variable denote the no. of times a total of 9 is obtained in two tosses of two dices. ∴ X can take values 0,1,2.
P(X = 0) = P(set getting a total of 9 in two tosses) = q
q = \(\frac{8}{9}\) × \(\frac{8}{9}\)
= \(\frac{64}{81}\)
P(X = 1) = P(getting a total of 9 one time in two draws)
= p q + q p
= \(\frac{1}{9}\) × \(\frac{8}{9}\) + \(\frac{8}{9}\) × \(\frac{1}{9}\)
= \(\frac{16}{81}\)
P(X = 2) = P(getting a total of 9, 2 times in two draws)
= p p
= \(\frac{1}{9}\) × \(\frac{1}{9}\)
= \(\frac{1}{81}\)
The table of values for competition of µ and ∑² is given as under:

x	P(X)	Xp(X)	Xpx
0	\(\frac{64}{81}\)	0	0
1	\(\frac{16}{81}\)	\(\frac{16}{81}\)	\(\frac{16}{81}\)
2	\(\frac{1}{81}\)	\(\frac{2}{81}\)	\(\frac{4}{81}\)
		∑ X P(X) = \(\frac{18}{81}\)	∑ X2 P(X) = \(\frac{20}{81}\)

∴ Mean = µ = ∑ X P(X)
= \(\frac{18}{81}\)
= \(\frac{2}{9}\)
and Variance = ∑²
= ∑ X² P(X) – µ²
= \(\frac{20}{81}\) – (\(\frac{2}{9}\))²
= \(\frac{20}{81}\) – \(\frac{4}{81}\)
= \(\frac{16}{81}\)

Question 10.
A die is tossed twice. A success is getting an odd number on a random toss. Find the variance of the number of successes.
Answer:
Let p = prob. of success = prob. of getting an odd number in a single toss of a diee = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Let X be the random variable denotes the no. of success in 2 tosses of single die. So X can take values 0,1,2.
P(X = 0) = q
q = \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\)
P(X = 1) = p q + q p
= \(\frac{1}{2}\) × \(\frac{1}{2}\) + \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1}{2}\)
P(X = 2) = p
p = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)
∴ Mean = 0 + \(\frac{1}{4}\) + 1 × \(\frac{1}{2}\) + 2 × \(\frac{1}{4}\)
= \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
and Variance = ∑X² P(X) – µ²
= 0² × \(\frac{1}{4}\) + 1² × \(\frac{1}{2}\) + 2² × \(\frac{1}{4}\) – 1²
= \(\frac{1}{2}\) + 1 – 1
= \(\frac{1}{2}\)
and
Variance = ∑X² P(X) – µ²
= 0² × \(\frac{1}{4}\) + 1² × \(\frac{1}{2}\) + 2² × \(\frac{1}{4}\) – 1²
= \(\frac{1}{2}\) + 1 – 1
= \(\frac{1}{2}\)