# Maharashtra Board 10th Class Maths Part 1 Practice Set 5.4 Solutions Chapter 5 Probability

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

## Practice Set 5.4 Algebra 10th Std Maths Part 1 Answers Chapter 5 Probability

Question 1.
If two coins are tossed, find the probability of the following events.
i. Getting at least one head.
Solution:
Sample space,
S = {HH, HT, TH, TT}
∴ n(S) = 4

i. Let A be the event of getting at least one head.
∴ A = {HT, TH, HH}
∴ n(A) = 3
∴ P(A) = $$\frac { n(A) }{ n(S) }$$
∴ P(A) = $$\frac { 3 }{ 4 }$$

ii. Let B be the event of getting no head.
∴ B = {TT}
∴ n(B) = 1
∴ P(B) = $$\frac { n(B) }{ n(S) }$$
∴ P(B) = $$\frac { 1 }{ 4 }$$
∴ P(A) = $$\frac { 3 }{ 4 }$$; P(B) = $$\frac { 1 }{ 4 }$$

Question 2.
If two dice are rolled simultaneously, find the probability of the following events.
i. The sum of the digits on the upper faces is at least 10.
ii. The sum of the digits on the upper faces is 33.
iii. The digit on the first die is greater than the digit on second die.
Solution:
Sample space,
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

i. Let A be the event that the sum of the digits on the upper faces is at least 10.
∴ A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 6
∴ P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 6 }{ 36 }$$
∴ P(A) = $$\frac { 1 }{ 6 }$$

ii. Let B be the event that the sum of the digits on the upper faces is 33.
The sum of the digits on the upper faces can be maximum 12.
∴ Event B is an impossible event.
∴ B = { }
∴ n(B) = 0
∴ P(B) = $$\frac { n(B) }{ n(S) }$$ = $$\frac { 0 }{ 36 }$$
∴ P(B) = 0

iii. Let C be the event that the digit on the first die is greater than the digit on the second die.
C = {(2, 1), (3, 1), (3,2), (4,1), (4,2), (4, 3), (5, 1), (5,2), (5,3), (5,4), (6,1), (6,2), (6, 3), (6, 4), (6, 5),
∴ n(C) = 15
∴ P(C) = $$\frac { n(c) }{ n(S) }$$ = $$\frac { 15 }{ 36 }$$
∴ P(C) = $$\frac { 5 }{ 12 }$$
∴ P(A) = $$\frac { 1 }{ 6 }$$ ; P(B) = 0; P(C) = $$\frac { 5 }{ 12 }$$

Question 3.
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn:
i. shows an even number.
ii. shows a number which is a multiple of 5.
Solution:
Sample space,
S = {1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15}
∴ n(S) = 15

i. Let A be the event that the ticket drawn shows an even number.
∴ A = {2, 4, 6, 8, 10, 12, 14}
∴ n(A) = 7
∴ P(A) = $$\frac { n(A) }{ n(S) }$$
∴ P(A) = $$\frac { 7 }{ 15 }$$

ii. Let B be the event that the ticket drawn shows a number which is a multiple of 5.
∴ B = {5, 10, 15}
∴ n(B) = 3
∴ P(B) = $$\frac { n(B) }{ n(S) }$$ = $$\frac { 3 }{ 15 }$$
∴ P(B) = $$\frac { 1 }{ 5 }$$
∴ P(A) = $$\frac { 7 }{ 15 }$$ ; P(B) = $$\frac { 1 }{ 5 }$$

Question 4.
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
i. an odd number?
ii. a multiple of 5?
Solution:
Sample space
(S) = {23, 25, 27, 29,
32, 35, 37, 39,
52, 53, 57, 59,
72, 73, 75, 79,
92, 93, 95, 97}
∴ n(S) = 20
i. Let A be the event that the number formed is an odd number.
∴ A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75,79,93,95,97}
∴ n(A) = 16
∴ P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 16 }{ 20 }$$
∴ P(A) = $$\frac { 4 }{ 5 }$$

ii. Let B be the event that the number formed is a multiple of 5.
∴ B = {25,35,75,95}
∴ n(B) = 4
∴ P(B) = $$\frac { n(B) }{ n(S) }$$ = $$\frac { 4 }{ 20 }$$
∴ P(B) = $$\frac { 1 }{ 5 }$$
∴ P(A) = $$\frac { 4 }{ 5 }$$ ; P(B) = $$\frac { 1 }{ 5 }$$

Question 5.
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is
i. an ace.
∴ P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 4 }{ 52 }$$
∴ P(A) = $$\frac { 1 }{ 13 }$$
∴ P(B) = $$\frac { n(B) }{ n(S) }$$ = $$\frac { 13 }{ 52 }$$
∴ P(B) = $$\frac { 1 }{ 4 }$$
∴ P(A) = $$\frac { 1 }{ 13 }$$ ; P(B) = $$\frac { 1 }{ 4 }$$