Selina Concise Class 10 Math Chapter 25 Exercise 25C Probability Solutions
Probability
Exercise: 25C
(Q1) A bag contains 3 red balls, 4 blue balls, and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:
(i) Yellow
(ii) Red
(iii) Not yellow
(iv) Neither yellow nor red
Solution:
A bag contains 3 red balls, 4 blue balls and 1 yellow ball.
∴ Total number of balls = Red + blue + yellow
= 3 + 4 + 1
= 8
∴ n(s) = 8
(i) Event A: Yellow balls {1}
n(A) = 1
Probability (Getting only a yellow ball) = n(A)/n(S)
P(A = 1/8
(ii) Even B: Red balls = {i.e. 3}
n(B) = 3
Probability (getting only red balls) = n(B)/n(S)
P(B) = 3/8
(iii) Event C: Not yellow balls = {i.e. red + blue balls = 3+4 = 7}
n(C) = 7
Probability (getting not yellow balls) = n(C)/n(s)
P(c) = 7/8
P(C) = 7/8
(iv) Event D: neither yellow nor red = {i.e. only blue balls = 4}
n(D) = 4
Probability (getting neither yellow nor red) = n(D)/n(S)
P(D) = n(D)/n(S) = 4/8
P(D) = ½
(Q2) A dice is thrown once. What is probability of getting a number:
(i) Greater than 2?
(ii) Less than or equal to 2?
Solution:
A dice is thrown once: i.e. {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) Event A: greater than 2 = {i.e. 3, 4, 5, 6}
n(A) = 4
Probability (getting only greater than 2 numbers) = n(A)/n(S)
P(A) = 4/6
P(A) = 2/3
(ii) Event B: less than or equal to 2 = {i.e. 1, 2}
n(B) = 2
Probability (getting less than or equal to numbers)
= n(B)/n(S)
P(B) = 2/6
p(B) = 1/3
(Q3) From a well shuffled deck of 52 cards one card is drawn, find the probability that the card drawn is:
(i) A face card
(ii) Not a face card
(iii) A queen of black card
(iv) A card with number —-
(v) A card with number less than 8
(vi) A card with number between 2 and 9
Solution:
A well – shuffled deck of 52 cards
Total number of cards is 52 n(S) = 52
(i) Event A: a face card = {i.e. total face cards = 12 out of total 52 cards)
n(A) = 12 (4 kings, 4 queens and 4 jacks)
Probability (getting a face cards) = n(A)/n(S)
P(A) = 12/52
P(A) = 3/13
(ii) Event B: Not a face card = {i.e. total deck of cards = 52 – face cards = 52 – 12 = 40}
n(S) = 40
Probability (getting not a face card) = n(B)/n(S)
P(B) = 40/52
P(B) = 10/13
(iii) Event C: a queen of black card = {i.e. queen (Spade, queen (club)}
N(c) = 2 cards
Probability (getting a queen of black cards) = n(C)/n(s)
p(C) = 2/52 cards
P(c) = 1/26
(iv) Event D: a card with number 5 or 6 {i.e. 5H, 5D, 5S, 5C, 6H, 6D, 65, 6C}
n(D) = 8
Probability (getting a card with number 5 or 6) = n(D)/n(S)
p(D) = 8/52
P(D) = 2/13
(v) Event E: a card with number less than 8 = {i.e. 2, 3, 4, 5, 6, 7}
{i.e. 6H cards, 6D cards, 65 cards, 6C cards}
∴ n(C) = 24
P(E) = n(E)/n(S)
= 24/52
P(E) = 6/13
(vi) Event F: a card with number between 2 and 9 = {3, 4, 5, 6, 7, 8}
{i.e. 6H cards, 6D cards, 6S cards, 6C cards}
∴ n(F) = 24 (6×4)
p(F) = n(F)/n(S)
= 24/52
P(F) = 6/13
(Q6) A man tosses two different coins (one of RS 2 and another of RS 5)
Simultaneously, what is the probability that he gets?
(i) At least one head
(ii) At most one head
Solution:
Two coins are tossed simultaneously.
{I.e. HH, HT, TH, TT}
n(S) = 4
(i) Event A: at least one head = {i.e. HH, HT, TH}
n(A) = 3
Probability (at least one head) = n(A)/n(S)
P(A) = ¾
(ii) Event B: at most one head = {i.e. TH, TT, HT}
n(B) = 3
Probability (getting at most one head) = n(B)/n(S)
P(B) = ¾
(Q7) A box contains 7 red balls, 8 green balls. A ball is drawn at random from the box. Find the probability that the ball is:
(i) White
(ii) Neither red nor white
Solution:
A box contains 7 re balls, 8 green balls and 5 white balls.
∴ Total number of balls = Red balls + green balls + white balls .
7 + 8 + 5 = 20 balls
n(S) = 20
(i) Event: A white balls = {i.e. Total number of white balls = 5}
n(A) = 5
Probability (getting only white balls) = n(A)/n(S)
p(A) = 5/20
P(A) = 1/4
(ii) Event B: neither Red nor white balls = {i.e. only green balls = 8}
n(B) = 8
Probability (getting green balls) = n(B)/n(S)
p(B) = 8/20
P(B) = 4/10 = 2/5
P(B) = 2/5
(Q8) All the three face cards of spades are removed from a well shuffled pack of 52 cards. A card is then drawn at random from the remaining pack, find the probability of getting:
(i) A black face card
(ii) A queen
(iii) A black card
Solution:
All the three face cards of spades are removed from a well shuffled pack of 52 cards
∴ Total number of cards are 52 according to question 3 face cards are removed from 52 cards.
∴ 52 – 3 = 49 cards
∴ n(S) = 49
(i) Even A: black face card = {i.e. total number of black face card are = 3}
n(A) = 3
Probability (getting a black face cards) = n(B)/n(S)
P(A) = 3/49
(ii) Event B: a queen = {i.e. total number of queen cards = 3}
n(B) = 3
Probability (getting a queen cards) = n(B)/n(S)
P(B) = 3/49
(iii) Event C: a black card = {i.e. the total number of black cards are 13 club + 10 spade = 23}
n(C) = 23
Probability (getting a black card) = n(c)/n(S)
P(C) = 23/49
(Q10) In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) It is acceptable to a trader who accepts only a good shirts?
(ii) It is acceptable to a trader who rejects only a shirts with major defects?
Solution
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects.
∴ Total number of shirts = 50
n(S) = 50
(i) Even A: a trader accepts any good so, the number of good shirts = 44
n(A) = 44
Probability (getting only good shirts) = n(A)/n(S)
P(A) = 44/50
P(A) = 22/25
(ii) Event B: A trader rejects major defects shirts and the total number of major defects shirts are 2.
Total number of shirts = 50 – major defects shirts
= 50 – 2
Accepting 9 shirts = 48
n(B) = 48
p(B) n(B)/n(S)
P(B) = 24/25
(Q11) Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8
(ii) 13
(iii) Less than or equal to 12
Solution:
Two dice are thrown at the same time
{i.e. (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (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, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
(i) Even A: A sum of the numbers is 8 = {(2, 6), (3, 5), (4, 6), (5, 3), (6,2)}
n(A) = 5
P(A) = n(A)/n(S)
P(A) = 5/36
(ii) Event B: A sum of the two numbers is 13 = there is no outcomes
∴ n(B) = 0
P(B) = n(B)/n(S)
P(B) = 0/36
P(B) = 0
(Q13) If p(E) = 0.59; find P(not E).
Solution: Given that,
P (E) = 0.59, find P (not E) =?
We know that,
P(E) + P (not E) = 1
=) 0.59 + P (not E) = 0
=) P (not E) = 1 – 0.59
=) P (not E) = 0.41
(Q15) The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?
Solution:
Given that, the probability that two do not have the same birthday is 0.897.
We know that,
P (do not have the same birthday) + P (have same birthday) = 1
0.897 + P(have same birthday) = 1
P (have same birthday) = 1 – 0.897
P (have same birthday) = 0.103
(Q16) A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn of the bag at random. What is the probability that the ball drawn will be:
(i) Not red?
(ii) Neither Red nor green?
(iii) White or green
Solution:
A bag contains 10 red balls, 16 white balls, and 8 green balls.
Total number of balls = red balls + white balls + green balls
= 10 + 16 + 8
= 34 balls
∴ n(S) = 34
(i) Event A: not red balls = {i.e. only white balls and green balls = 16 + 8 = 24}
∴ n(A) = 24
P(A) = n(A)/n(S)
= 24/34
P(A) = 12/17
(ii) Event B: neither red no green = {i.e. only balls = 16}
n(B) = 16
P(B) = n(B)/n(S)
p(B) = 16/34
p(B) = 8/17
(iii) Event C: white or green = {I.e. white or green balls = 16+8 = 24}
n(C) = 24
P(C) = n(C)/n(S)
= 24/34
P(C) = 12/7
(Q17) A bag contains twenty RS 5 coins, fifty RS 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
(i) Will be a Re 1 coin?
(ii) Will not be a RS 2 coin?
(iii) Will neither be a RS 5 coin nor be a Re 1 coin?
Solution:
A bag contains twenty RS 5 coins, fifty RS 2 coins and thirty Re 1 coins.
∴ Total number of coins = Twenty RS 5 coins + fifty RS 2 coins + thirty Re 1 coins
20 + 50 + 30
= 100
∴ n(S) = 100
(i) Event A: Will be a Re 1 coin = {i.e. Thirty Re 1 coin = 30}
∴ n(A) = 30
∴ p(A) = n(A)/n(S)
= 30/100
P(A) = 3/10
(ii) Event B: Will not be a RS 2 coin = {I.e. only 1 or RS 5 coins = 30 + 20 = 50}
n(B) = 50
P(B) = n(B)/n(S)
= 50/100
P(B) = ½
(iii) Event C: will neither be a RS 5 coin nor be a Re 1 coin = {i.e. only RS 2 coins = 50}
n(C) = 50
p(C) = n(C)/n(S)
= 50/100
P(C) = ½
(Q18) A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6 7, 8, 9, 10, 11, 12; as shown below.
If the outcomes are equally likely, find the probability that the pointer will point at:
(i) 6
(ii) An even number
(iii) A number less than or equal to 9
(iv) A prime number
(v) A number between 3 and 11
(vi) A number greater than 8
Solution:
Total number = 12
{i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
∴ n(S) = 12
(i) Event A: 6 = {i.e. only one numbers}
n(A) = 1
P(A) = n(A)/n(S)
P(A) = 1/12 {The pointer will point at 6}
(ii) Event B: an even number = {i.e. 2, 4, 6, 8, 10, 12}
n(B) = 6
P(The pointer will be at an even number) = n(A)/n(S)
P(B) = 6/12
P(B) = ½
(iii) Event C: a prime number = {i.e. 2, 3, 5, 7, 11}
n(C) = 5
P(The pointer will be at a prime number) = n(C)/n(S)
p(C) = 5/12
(iv) Event D: A number greater than 8 = {i.e. 9, 10, 11, 12}
n(D) = 4
P(The pointer will be at a number greater than 8) = n(D)/n(S)
p(D) = 4/12
p(D) = 1/3
(v) Event E: a number less than or equal to 9 = {i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9}
n(E) = 9
∴ P (The pointer will be at a umber less than or equal to 9) = n(E)/n(S)
P(e) = 9/12
P(E) = 3/4
