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 = \(\frac{n(\mathrm{D})}{n(\mathrm{~S})}\) = \(\frac{4}{16}\) = \(\frac{1}{4}\)