Math Olympiad Test: Probability- 3 - Free MCQ with solutions Class 10

Leap year contains 366 days.
⇒ 52 weeks + 2 days
52 weeks contain 52 Fridays.
We will get 53 Fridays if one of the remaining two days is a Friday.
Total possibilities for two days are:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
There are 7 possibilities and out of these 2 are favourable cases.
Required probability = 2/7

Total number of outcomes when two dice are thrown = 36.
Let A be the event of getting a number always greater than 4 on second dice.
∴ A = {(1,5), (1,6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6), 
(5, 5), (5, 6), (6, 5), (6, 6)}
∴ Number of possible outcomes = 12
∴ P(A) = 12/36 = 1/3

Total 13 cards are present in suit of club if 3 cards are removed, then 10 cards of clubs are remaining.
Total cards remaining = 49
∴ Probability of getting a club = 10/49

Outcomes are 13, 14, 15,.......,60.
Total number of possible outcomes = 60 - 12 = 48
(i) The numbers divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60.
Thus, the number of numbers divisible by 5 = 10 
Required probability = 10/48 = 5/24
(ii) Perfect square numbers are 16, 25, 36, 49
Thus, the number of perfect square number = 4 
Required probability =  4/48 = 1/12

Total number of marbles = 3 + 4 + 2 = 9 
No. of green and blue marbles = 3 + 4 = 7.
∴ Probability of not getting an orange marble = 7/9

Total cards = 5
(i) Number of queens = 1
⇒ P(queen) = 1/5
(ii) If the queen is drawn and put aside, then the remaining cards = 5 − 1 = 4
(a) Number of aces = 1
⇒ P(ace) = 1/4
(b) Number of queens = 0
⇒ P(queen) = 0

Total number of cards = 52
Total number of aces present in a deck of cards = 4
∴ Probability of drawing an ace = 4/52 = 1/13

Total number of outcomes, when two dice are thrown = 6 x 6 = 36
Total number of doublets present = (1,1), (2, 2), (3, 3), (4, 4), (5, 5), (6,6)
For a total of 4, pairs can be (1,3), (3, 1), (2, 2).
= 8 [since (2, 2) is present in both the cases],
∴ Required probability = 8/36 = 2/9

Total number of outcomes = 250
Number of favourable outcomes = 5
∴ Probability that Kunal wins the prize = 5/250 = 1/50

Out of 600 electric bulbs one bulb can be chosen in 600 ways.
∴ Total number of elementary events = 600
There are 588 (= 600 - 12) non-defective bulbs out of which one bulb can be chosen in 588 ways.
∴ Favorable number of elementary events = 588
Hence. P (getting a non-defective bulb) = 588/600
= 49/50
= 0.98