Selina Concise Class 10 Math Chapter 25 Exercise 25C Probability Solutions

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


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