Maharashtra Board Class 10 Math Part – 1 Solution Chapter 5 Practice Set 5.4 – Probability
Balbharati Maharashtra Board Class 10 Math Part – 1 Solution Chapter 5: Probability. Marathi or English Medium Students of Class 10 get here Probability full Exercise Solution.
Std | Maharashtra Class 10 |
Subject | Math Part 1 Solution |
Chapter | Probability |
Practice Set | 5.4 |
Practice Set 5.4
(1) If two coins are tossed, find the probability of the following events.
(1) Getting at least one head.
(2) Getting no head
Solution:
Two coin is tossed.
S = {HT, HH, TH, TT}
n (s) = 4
(1) Event A: getting at least one head.
A = {HT, HH, TH}
n (A) = 3
∴ P (A) = n(A) / n(S) = 3/4
(2) Event B: getting no head
B = {TT}
n (B) = 1
∴ P (B) = n (B)/n(S) = 1/4
(2) If two dice are rolled simultaneously, find the probability of the following events.
(1) The sum of the digits on the upper faces is at least 10.
(2) The sum of the digits on the upper faces is 33.
(3) The digit on the first die is greater than the digit on second die.
Solution:
Two die is rolled
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
(1) Event A: 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) = n(A)/n(S) = 6/36 = 1/6
(2) Event B: Sum of the digits on upper face is 33
B = {not possible}
n (B) = 0
∴ P (3) = n(B)/A(s) = 0/36 = 0
Event C: 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) = n (c)/n (s) = 15/36 = 5/12
(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 –
(1) Shows an even number.
(2) Shows a number which is a multiple of 5.
Solution:
Total no. of tickets are 15 with each different numbers.
Let, S be the sample space,
∴ S = {1, 2, 3, 5, 6 7, 8, 9, 10, 11 12, 13, 14, 15}
n (s) = 15
(1) Event A: Shows an event number.
A = {2, 4, 6, 8, 10, 12, 14}
n (A) = 7
∴ P (A) = n (A)/n (s) = 7/15
(2) Event B: Shows a number which is a multiple of 5.
B = {5, 10, 15}
n (B) = 3
∴ P (B) = n (B)/n (s) = 3/15 = 1/5
(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
(1) an odd number?
(2) a multiple of 5?
Solution:
Two digit number is formed with 2, 3, 5, 7, 9 without – repetition.
∴ S = {28, 25, 27, 29, 32, 35, 37, 39, 52, 53, 57, 59, 72, 73, 75, 79, 92, 93, 95, 97}
n (s) = 20
(1) Event A: An odd number.
A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 77, 79, 93, 95, 97}
n (A) = 16
∴ P (A) = n (A)/n (s) = 16/20 = 4/5
(2) Event B: Multiple 75.
B = {25, 35, 75, 95}
n (B) = 4
∴ P (B) = n (B)/n (s) = 4/20 = 1/5.
(5) A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is –
(1) An ace.
(2) A spade.
Solution:
A card is drawn from a deck of cards.
∴ Let, S be the sample space
∴ n (s) = 52
(1) event A – an ace
n (A) = 4 [∵ There are 4 aces of each suit]
∴ P (A) = n (A)/n (s) = 4/52 – 1/13
(2) event B:- a spade
A (B) = 13 [∵ there are 13 spades is a deck of cards]
∴ P (B) = n(B)/n(s) = 13/52 = 1/4
Here is your solution of Maharashtra Board Class 10 Math Part – 1 Chapter 5 Probability Practice Set 5.4
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