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Math Olympiad Test: Probability- 3 - Class 10 MCQ


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10 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Probability- 3

Math Olympiad Test: Probability- 3 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Probability- 3 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Probability- 3 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Probability- 3 below.
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Math Olympiad Test: Probability- 3 - Question 1

Find the probability of getting 53 Fridays in a leap year.

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 1

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

Math Olympiad Test: Probability- 3 - Question 2

When two dice are thrown, the probability of getting a number always greater than 4 on the second dice is _____.

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 2

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

Math Olympiad Test: Probability- 3 - Question 3

The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well-shuffled. One card is selected from the remaining cards. The probability of getting a club is ______.

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 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

Math Olympiad Test: Probability- 3 - Question 4

Cards marked wit h numbers 13, 14, 15,.......,60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the drawn card is
(i) divisible by 5.
(ii) a number which is a perfect square.

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 4

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

Math Olympiad Test: Probability- 3 - Question 5

A bag contains three green, four blue and two orange marbles. If a marble is picked at random, then the probability that it is not an orange marble, is ______.

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 5

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

Math Olympiad Test: Probability- 3 - Question 6

Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. If the queen is drawn and put aside, one card is then picked up at random. what is the probability that the second card picked up is
(i) a king and
(ii) a queen?

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 6

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

Math Olympiad Test: Probability- 3 - Question 7

One card is drawn from a well-shuffled deck of 52 cards. The probability of drawing an ace is _____.

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 7

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

Math Olympiad Test: Probability- 3 - Question 8

Two dice are thrown simultaneously. The probability of getting a doublet or a total of 4 is _____.

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 8

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

Math Olympiad Test: Probability- 3 - Question 9

250 lottery tickets were sold and there are 5 prizes on these tickets. If Kunal has purchased one lottery ticket, what is the probability that he wins a prize?

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 9

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

Math Olympiad Test: Probability- 3 - Question 10

It is know that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?

Detailed Solution for Math Olympiad Test: Probability- 3 - Question 10

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

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