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Test Level 2: Probability - 1 - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 2: Probability - 1

Test Level 2: Probability - 1 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 2: Probability - 1 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 2: Probability - 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 2: Probability - 1 below.
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Test Level 2: Probability - 1 - Question 1

There are 5 Engineering books, 4 Mathematics books and 2 Physics books on a shelf, placed randomly. Find the probability that all the books of each kind are placed together.

Detailed Solution for Test Level 2: Probability - 1 - Question 1

Treat the 5 Engineering books as 1 unit
Treat the 4 Mathematics books as 1 unit
Treat the 2 Physics books as 1 unit
Then there are 3 objects, which can be arranged in 3! ways, but amongst each subject, the books can be arranged in 5!4!2! ways, so
Required Probability


Test Level 2: Probability - 1 - Question 2

Tickets numbered from 1 to 25 are mixed up together and a ticket is drawn at random. What is the probability that the drawn ticket contains a odd number?

Detailed Solution for Test Level 2: Probability - 1 - Question 2

Total odd numbers between 1 to 25 = 13
Hence, probability = 13/25

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Test Level 2: Probability - 1 - Question 3

A pair of dice is thrown. Find the probability of the numbers appearing whose sum is greater than or equal to 10.

Detailed Solution for Test Level 2: Probability - 1 - Question 3

Total number of possible cases = 36
Favourable events = (4, 6) (5, 5) (6, 4) (5, 6) (6, 5) (6, 6) = 6
Required probability = 6/36 = 1/6

Test Level 2: Probability - 1 - Question 4

One hundred identical coins, each with probability 'p' of showing up heads, are tossed once. If the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, find the value of p.

Detailed Solution for Test Level 2: Probability - 1 - Question 4

Suppose, X denotes the number of coins showing heads up.
Then, X is a binomial variate with n = 100 and probability of success p.
We have, P (X = 51) = P (X = 50)
100C51 (p)51 (1 - p)49 = 100C50 (p)50 (1 - p)50   

50p = 51 - 51p
101p = 51
p = 51/101

Test Level 2: Probability - 1 - Question 5

In a box, there are 5 Nestle and 18 Dairy Milk chocolates. Mohan picks up a chocolate without looking at it. What is the probability that it will be a Nestle chocolate?

Detailed Solution for Test Level 2: Probability - 1 - Question 5

Number of Nestle chocolates = 5
Number of Dairy Milk chocolates = 18
Total number of chocolates = 18 + 5 = 23
Probability that a picked up chocolate is Nestle = 5/23

Test Level 2: Probability - 1 - Question 6

A trader saves 10% in buying goods from a manufacturer in place of that from a wholesaler and sells at 10% above the wholesale price. What percent does he gain overall?

Detailed Solution for Test Level 2: Probability - 1 - Question 6

Suppose wholesale price = Rs. 100
Since the trader buys at Rs. 90 and sells at Rs. 110,
Therefore, the required profit percentage = [(110 - 90)/90] × 100 = 22.22%

Test Level 2: Probability - 1 - Question 7

Five boys and 3 girls are to be seated on chairs arranged in a row. If the arrangement is made at random, find the probability that no two girls will be seated next to each other.

Detailed Solution for Test Level 2: Probability - 1 - Question 7

Five boys and three girls can be seated on 8 chairs in 8! different ways.
_ B _ B _ B _ B _ B _
As shown above, 5 boys can be seated at 5 places in 5! ways. After this, 3 girls can be seated in the remaining 6 places in 6P3 ways.
Favourable outcomes = 5! × 6P3
Probability = (5!/3!) × (6!/8!) = 5/14

Test Level 2: Probability - 1 - Question 8

In a class, 30% of the students passed in English, 20% of the students passed in Hindi and 10% passed in both. If a student is selected at random, what is the probability that he passed in English or Hindi?  

Detailed Solution for Test Level 2: Probability - 1 - Question 8

P(E) = 30/100 = 3/10, P(H) = 20/100 = 1/5, P(E ∩ H) = 10/100 = 1/10.
P(English or Hindi) = P(E ∪ H) = P(E) + P(H) – P(E ∩ H) = 4/10 = 2/5.

Test Level 2: Probability - 1 - Question 9

A sample space consists of numbers from 0 to 20. What is the probability that an element of the sample space is an even number, excluding numbers 0 and 20?  

Detailed Solution for Test Level 2: Probability - 1 - Question 9

Sample space is 21.
Even numbers between 0 and 20 are {2, 4, 6, 8, 10, 12, 14, 16, 18}
Number of favourable cases = 9
Hence, the probability is 9/21 = 3/7.

Test Level 2: Probability - 1 - Question 10

Two dices are thrown simultaneously. What is the probability of getting two numbers whose sum is odd?  

Detailed Solution for Test Level 2: Probability - 1 - Question 10

S = sample space. n(S) = 36.
E = event of getting two numbers whose sum is odd = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}.
n(E) = 18.

P(E) = n(E)/n(S) = 18/36 = 1/2.

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