# Class 10 RD Sharma Solutions – Chapter 13 Probability – Exercise 13.1 | Set 2

• Last Updated : 08 Oct, 2021

### (i) What is the probability that the card is a queen?

Solution:

Total cards = 5

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Total queen = 1

Number of favorable outcomes = 1

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting cards which is a queen = 1/5

### (a) ace?

Solution:

Total cards after king = 4

Number of favorable outcomes = 1

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting an ace card = 1/4

### (b) king?

Solution:

Number of favorable outcomes = 0

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting a king = 0

### (i) Red

Solution:

Total number of balls = 3 + 5 = 8

Total red balls = 3

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of drawing a red ball = 3/8

### (ii) Back

Solution:

Total black ball = 5

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of drawing a black ball = 5/8

### (i) 10?

Solution:

Total numbers on the spin = 12

Favorable outcomes = 1

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting a 10 = 1/12

### (ii) an odd number?

Solution:

Favorable outcomes are 1, 3, 5, 7, 9, and 11

Favorable outcomes = 6

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting a prime number = 6/12 = 1/2

### (iii) a number which is multiple of 3?

Solution:

Favorable outcomes are 3, 6, 9, and 12

Favorable outcomes = 4

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting multiple of 3 = 4/12 = 1/3

### (iv) an even number?

Solution:

Favorable outcomes are 2, 4, 6, 8, 10, and 12

Favorable outcomes = 6

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting an even number = 6/12 = 1/2

### (i) The name of a girl

Solution:

Total number of students in the class = 18 + 16 = 34

Favorable cases = 18

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting a name of girl on the card = 18/34 = 9/17

### (ii) The name of a boy?

Solution:

Favorable cases = 16

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of getting a name of boy on the card = 16/34 = 8/17

### Question 23. Why is tossing a coin considered to be a fair way of deciding which team should choose ends in a game of cricket?

Solution:

Possible outcomes while tossing a coin = 2  (1 head or 1 tail)

Probability = Number of favorable outcomes/ Total number of outcomes

P(getting tail) = 1/2

As the probability of both the events are equal

Therefore, tossing a coin is considered to be a fair way of deciding which team should choose ends in a game of cricket.

### Question 24. What is the probability that a number selected at random from the number 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 will be their average?

Solution:

Total number of possible outcomes = 10

Average of the outcomes = (1 + 2 + 2+3+3+3+4+4+4+4) / 10

= 30/10

= 3

Let E be the event of getting 3.

Number of favorable outcomes = 3

P(E) = Number of favorable outcomes/ Total number of outcomes

P(E) = 3/10

Therefore, the probability that a number selected at random will be the average is 3/10.

### Question 25. There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.

Solution:

Total number of possible outcomes are 30 {1, 2, 3, … 30}

Let E = event of getting a number that is divisible by 3

Number of favorable outcomes = 10{3, 6, 9, 12, 15, 18, 21, 24, 27, 30}

Probability, P(E) = Number of favorable outcomes/ Total number of outcomes

P(E) = 10/30

= 1/3

P(not E) = 1- 1/3

= 2/3

Therefore, the probability that the number on the selected card is not divisible by 3 = 2/3

### Question 26. A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is

Solution:

(i) red or white

Total number of possible outcomes = 20 (5 red, 8 white & 7 black}

Number of favorable outcomes = 13 (5 red + 8 white)

Probability = Number of favorable outcomes/ Total number of outcomes

Therefore, the probability of red or white ball = 13/20

(ii) not black

No. of favorable outcomes =7 (7 black balls)

Probability = Number of favorable outcomes/ Total number of outcomes

Probability of black ball = 7/20

= 7/20

P (not E) = 1-7/20

= 13/20

(iii) neither white nor black.

Number of favorable outcomes = 20 – 8 – 7 = 5(total balls – no. of white balls – no. of black balls)

Probability = Number of favorable outcomes/ Total number of outcomes

P(E) = 5/20 = 1/4

### Question 27. Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.

Solution:

Total no. of possible outcomes = 25 {1, 2, 3…. 25}

Favorable outcomes are 2, 3, 5, 7, 11, 13, 17, 19, 23

Number. of favorable outcomes = 9

Probability= Number of favorable outcomes/ Total number of outcomes

P(E) = 9/25

P(not E) = 1-9/25

= 16/25

Therefore, the probability of selecting a number which is not prime is 16/25.

### (i) Red or white

Solution:

Total number of balls = 8 + 6 + 4 = 18

Total outcomes =18

Favorable outcomes = 14 (8 red balls + 6 white balls)

Probability = Number of favorable outcomes/ Total number of outcomes

= 14/18

= 7/9

### (ii) Not black

Solution:

Let E be event of getting a black ball

Number of favorable outcomes = 4

P(E) = 4/18

P(E) = 2/9

P(not E) = 1-2/9

= 7/9

Therefore, probability of not a black ball is 7/9

### (iii) Neither white nor black

Solution:

Let E be event of getting neither a white nor a black ball

Favorable outcomes = 18 – 6 – 4

= 8

Probability= Number of favorable outcomes/ Total number of outcomes

P(E) = 8/18 = 4/9

### (i) Prime number

Solution:

Total outcomes = 35

Favorable outcomes =11 {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31)

Probability = Number of favorable outcomes/ Total number of outcomes

= 11/35

Therefore, the probability of prime number = 1/22

### (ii) Multiple of 7

Solution:

Favorable outcomes = 5 {7, 14, 21, 28, 35}

Probability = Number of favorable outcomes/ Total number of outcomes

= 5/35

= 1/7

Therefore, the probability of multiple of 7 = 1/22

### (iii) Multiple of 3 or 5

Solution:

Favorable outcomes = 16 {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20, 25, 35}

Probability = Number of favorable outcomes/ Total number of outcomes

= 16/35

Therefore, the probability of multiple of 3 or 5= 1/22

### (i) a black queen

Solution:

Total cards = 52

All jacks, queens & kings, ace of red color are removed.

Total outcomes = 52 – 2 – 2 – 2 – 2 = 44 (remaining cards)

Favorable outcomes = 2 (queen of spade & club)

Probability = Number of favorable outcomes/ Total number of outcomes

= 2/44

=1/22

Therefore, the probability of black queen = 1/22

### (ii) a red card

Solution:

Favorable outcomes = 26 – 8

= 18

Probability = Number of favorable outcomes/ Total number of outcomes

= 18/44

= 9/22

Therefore, the probability of red card = 1/22

### (iii) a black jack

Solution:

Favorable outcomes = 2 (jack of club & spade)

Probability = Number of favorable outcomes/ Total number of outcomes

= 2/44

= 1/22

Therefore, the probability of black jack = 1/22

### (iv) a picture card (Jacks, queens and kings are picture cards)

Solution:

Favorable outcomes = 6 (2 jacks, 2 kings & 2 queens of black color)

Probability = Number of favorable outcomes/ Total number of outcomes

= 6/44

= 3/22

Therefore, the probability of picture card = 3/22

### (i) an orange-flavoured candy

Solution:

Taking out orange flavored candy is an impossible event because all lemon candies are there.

Therefore, probability = 0

### (ii) a lemon flavored candy

Solution:

As the bag contains all lemon candies, taking out lemon candy is sure event

Therefore, probability = 1

### Question 32. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

E = event of 2 students not having same birthday

P(E) = 0.992

P(not E) = 1 – 0.992

= 0.008

Therefore, the probability that the 2 students have the same birthday is 0.008

### (i) red

Solution:

Total outcomes = 8

Favorable outcomes = 3

Probability = Number of favorable outcomes/ Total number of outcomes

= 3/8

Therefore, the probability of red ball = 3/8

### (ii) not red

Solution:

P(not red) = 1- P(red)

= 1-3/8

=5/8

Therefore, the probability of not red ball = 5/8

### (i) red

Solution:

Total outcomes = 17

Favorable outcomes = 5

Probability = Number of favorable outcomes/ Total number of outcomes

= 5/17

Therefore, the probability of red marble = 5/17

### (ii) not green

Favorable outcomes = 4

Probability = Number of favorable outcomes/ Total number of outcomes

= 4/17

P(not green) = 1 – 4/17

= 13/17
Therefore, the probability that the marble taken out is not green is 13/17.

### (i) She will buy it

Solution:

Good pens = 144 – 20 = 124

Defective pens = 20

Total outcomes =144

Favorable outcomes = 124

Probability = Number of favorable outcomes/ Total number of outcomes

= 124/144

= 31/36

Therefore, the probability that she will buy = 31/36

### (ii) She will not buy it

Solution:

=5/36

Therefore, the probability that she will not buy = 5/36

### Question 36. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at pen and tell whether it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is good one.

Solution:

Good pens = 132

Defective pens = 12

Total outcomes = 144

Favorable outcomes = 132

Probability = Number of favorable outcomes/ Total number of outcomes

= 132/144

= 11/12

Therefore, probability of good pen = 11/12

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