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