**Selina Concise Class 10 Math Chapter 25 Exercise 25B Probability Solutions**

**Probability**

__Exercise: 25B__

__Exercise: 25B__

**(Q1) Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random find the probability that the card selected will be:**

**(i) An even number **

**(ii) A multiple of 3 **

**(iii) An even number and a multiple of 3 **

**(iv) An even number or a multiple of 3 **

**Solution: **

Given that, there are total nine cards are numbered 2 to 10

n(S) = 9

(i) Event A: an even number = {2, 4, 6, 8, 10} = 5

n(A) = 5

P(A) = n(A)/n(S)

= 5/9

P(A) = 5/9

(ii) Event B a multiple of 3 = {3, 6, 9} = 3

n(B) = 3

Probability (Getting only multiple of 3) = n(B)/n(S)

P(B) = 3/9 = 1/3

P(B) = 1/3

(iii) Event C: even number and multiple of 3 = {6} = 1

{6 is even number also it’s multiple of 3}

n(C) = 1

P(C) = n(C)/n(S) = 1/9

(iv) Event E: an even number or a multiple of 3 = {2, 3, 4, 6, 8, 9, 10}

2, 3, 4, 6 8, 9 and 10 these {Number are even or also multiple of}

n(E) = 7

P(E) = n(E)/n(S)

P(E) = 7/9

** **

**(Q2) Hundred identical cards are numbered from 1 to 100. The cards the cards are well shuffled and then a card is drawn find the probability that the number on card drawn is: **

**(i) A multiple of 5 **

**(ii) Multiple of 6 **

**(iii) Between 40 and 60 **

**(iv) Greater than 85 **

**(v) Less than 48 **

**Solution: **

Given that, three are 100 identical cards are numbered from 1 to 100.

∴ n(S) = 100

(i) Event A: a multiple of 5 = {i.e. 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100}

∴ n(A) = 20

Probability (Getting a multiple of 5 numbers) = n(A)/n(S)

P(A) = n(A)/n(S)

= 20/100

P(A) = 1/5

(ii) Event B: a multiple of 6 = {i.e. 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96}

n(B) = 16

Probability (getting a multiple of 6 numbers) = n(B)/n(S)

p(B) = 16/100

=) P(B) = 4/25

(iii) Event C: Between 40 and 60 = {i.e. 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59}

These numbers between 40 and 60

n(C) = 19

P(C) = n(C)/n(S) = 19/100

(iv) Event E: Greater than 85 = {I.e. 86, 87, 88, 89, 90, 91, 92, 93, 94, 95,96, 97, 98, 99, 100}

These numbers are greater than 85 only.

n(E) = 15

P(E) = n(E)/n(S)

P(E) = 15/100

P(E) = 3/20

(v) Event F: less than 48 = {i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, — 45, 46, 47

n(F) = 47

P (getting less than 48 numbers) = n(F)/n(S)

P(F) = n(F)/n(S)

P(F) = 47/100

** **

**(Q3) From 25 identical cards numbered 1, 2, 3, 4, 5, —— 24, 25: **

**One card is drawn at random find the probability that the number on the card drawn is a multiple of: **

**(i) 3 **

**(ii) 5 **

**(iii) 3 and 5 **

**(iv) 3 or 5 **

**Solution: **

Given that, there are 25 identical cards numbered 1, 2, 3, 4, 5 —– 24, 25

∴ n(S) = 25

(i) Event A: multiple of 3 = {i.e. 3, 6, 9, 12, 15, 18, 21, 24}

n(A) = 8

Probability (Getting a multiple of 3 numbers only) = n(A)/n(S)

P(A) = n(A)/n(S)

= 8/25

P(A) = 8/25

(ii) Event B: A multiple of 5 = {i.e. 5, 10, 15, 20, 25}

n(B) = 5

Probability (getting a multiple of 5 numbers only) = n(B)/n(S)

p(B) = n(B)/n(S)

P(B) = 5/25

P(B) = 1/5

(iii) Event C: a multiple of 3 and 5 = {i.e. 15}

n(C) = 1

Probability (Getting only a multiple of 3 and 5) = n(C)/n(S)

p(C) = 1/25

P(C) = 1/25

(iv) Event E: a multiple of 3 or 5 = {i.e. 3, 5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25}

n(E) = 12

Probability (getting only a multiple of 3 or 5) = n(E)/n(S)

P(E) = 12/25

** **

**(Q4) A die is thrown once. Find the probability of getting a number: **

**(i) Less than 3 **

**(ii) Greater than or equal to 4 **

**(iii) Less than 8 **

**(iv) Greater than 6**

**Solution: **

A die is thrown once

{I.e. 1, 2, 3, 4, 5, 6}

n(S) = 6

(i) Event A: Less than 3 = (i.e. 1, 2}

n(A) = 2

Probability (Getting only less than 3 numbers) = n(A)/n(S)

P(A) = n(A)/n(S)

= 2/6

P(A) = 1/3

(ii) Event B: greater than or equal to 4 = {i.e. 4, 5, 6}

n(B) = 3

Probability (Getting greater than or equal to 4 numbers only) = n(B)/n(S)

P(B) = 3/6

P(B) = 1/2

(iii) Event C: Less than 8 = {i.e. 1, 2, 3, 4, 5, 6}

n(C) = 6

On a dice numbers only 1 to 6.

So, n(C) = 6

Probability (getting less than 8 numbers) = n(C)/n(S)

P(C) = 6/6

P(C) = 1

(iv) Event E: greater than 6 = {i.e. 0}

because on dice there are only 1 to 6 numbers.

∴ n(E) = 0

P(E) = 0

P(E) = n(E)/n(S)

= 0/6

P(E) = 0

** **

**(Q5) A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8? **

**Solution: **

Given that, A book contains 85 pages.

∴ Total number of pages are 85.

∴ n(S) = 85

Sum of the digits on the page is 8 =

{i.e. 8, 9+8 = 17, 18+8 = 26, 27+8 = 35, 36+8 = 44, 45+8 = 53, 54+8 = 62, 63+8 = 71, 72+8 = 80}

{8, 17, 26, 35, 44, 53, 62, 71, 80}

n(A) = 9

Probability (Getting sum up to 8) = n(A)/n(S)

P(A) = 9/85

** **

**(Q6) 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: **

Given that a pair of dice is thrown.

A dice thrown at once is {1, 2, 3, 4, 5, 6}

A pair of dice is thrown is –

{(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

Probability (getting sum of 10 or more with 5) = n(A)/n(S)

Probability ((5, 5), (5, 6)) = n(A)/n(S)

n(A) = 2

P(A) = n(A)/n(S)

= 2/36

P(A) = 1/18

** **

**(Q7) If two coins are tossed once, what is the probability of getting: **

**(i) Both heads **

**(ii) At least one head **

**(iii) Both heads or both tails **

**Solution: **

If two coins are tossed together then the possible outcomes = {HH, TH, TT, HT}

∴ n(S) = 4

(i) Even A: both heads = {i.e. HH}

n(A) = 1

P(A) = n(A)/n(S)

P(A) = 1/4

(ii) Event B: at least one head = {i.e. HH, HT, TH}

n(B) = 3

∴ P(B) = n(B)/n(S)

P(B) = 3/4

(iii) Event C: Both heads or both tails = {i.e. HH, TT}

n(C) = 2

P(C) = n(C)/n(S)

P(C) = 2/4

P(C) = ½