CAT Exam  >  CAT Tests  >  Level-wise Tests for CAT  >  Test Level 3: Probability - 2 - CAT MCQ

Test Level 3: Probability - 2 - CAT MCQ


Test Description

10 Questions MCQ Test Level-wise Tests for CAT - Test Level 3: Probability - 2

Test Level 3: Probability - 2 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 3: Probability - 2 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 3: Probability - 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 3: Probability - 2 below.
Solutions of Test Level 3: Probability - 2 questions in English are available as part of our Level-wise Tests for CAT for CAT & Test Level 3: Probability - 2 solutions in Hindi for Level-wise Tests for CAT course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt Test Level 3: Probability - 2 | 10 questions in 20 minutes | Mock test for CAT preparation | Free important questions MCQ to study Level-wise Tests for CAT for CAT Exam | Download free PDF with solutions
Test Level 3: Probability - 2 - Question 1

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets a 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability of getting a 1, 2, 3 or 4 on the die?

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

E1 = Getting 5 or 6 in a single throw of a die.
E2 = Getting 1, 2, 3 or 4 in a single throw of a die.
A = Getting exactly one head.
P(E1) = 2/6 = 1/3.
P(E2) = 4/6 = 2/3.
P(A/E1) = Probability of getting exactly one head when a coin is tossed three times = 3C1 × (1/2)1 × (1/2)2 = 3/8.
P(A/E2) = Probability of getting exactly one head when a coin is tossed once only = 1/2.
Now, Required probability = P(E2/A) 

Test Level 3: Probability - 2 - Question 2

Two balls are drawn at random from a bag containing 3 white, 3 red, 4 green and 4 black balls one by one without replacement. Find the probability that both the balls are of different colours.

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

Probability that the 1st ball is white and the 2nd ball is other than white =

Probability that the 1st ball is red and the 2nd ball is other than red 

Probability that the 1st ball is green and the 2nd ball is other than green

Probability that the 1st ball is black and the 2nd ball is other than black 

Probability that both the balls are of different colours 

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test Level 3: Probability - 2 - Question 3

The probabilities that three men hit a target are 1/6, 1/4 and 1/3. Each man shoots once at the target. What is the probability that exactly one of them hits the target?

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

Probabilities that 3 men hit a target are 1/6 , 1/4 and 1/3 and
Probabilities that the same 3 men do not hit the target are 5/6, 3/4 and 2/3.
Let,
Event (A) = Exactly one of them hits the target 1st person or 2nd person or 3rd person

Test Level 3: Probability - 2 - Question 4

Three bags contain 18 balls of red and blue colours. The first bag contains 3 red and 3 blue balls. The second bag contains 4 red and 2 blue balls and the third bag contains 2 red and 4 blue balls. One ball is taken out at random from each bag. The probability of getting 2 red balls and 1 blue ball is

Detailed Solution for Test Level 3: Probability - 2 - Question 4

The possible cases of taking out 2 red and 1 blue balls are:


Test Level 3: Probability - 2 - Question 5

In a tailor shop, tailor A makes 500 shirts, tailor B makes 1200 shirts and tailor C makes 1600 shirts. Out of these, 200, 300 and 300 shirts are defective, respectively. One shirt is drawn at random and if it is found to be defective, then the probability that is made by tailor C is

Detailed Solution for Test Level 3: Probability - 2 - Question 5

Let Z1 = Event that the shirt is made by tailor A;
Z2 = Event that the shirt is made by tailor B;
Z3 = Event that the shirt is made by tailor C; and
z = Event that the shirt is defective.        
Total number of shirts made by these three tailors = 500 + 1200 + 1600 = 3300
P(Z1) = Probability that the shirt is made by tailor A = 500/3300 = 5/33
P(Z2) = Probability that the shirt is made by tailor B = 1200/3300 = 4/11
P(Z3) = Probability that the shirt is made by tailor C = 1600/3300 = 16/33
Total number of defective shirts made by three tailors = 200 + 300 + 300 = 800
Given, P(z/Z1) = Probability that the shirt is defective given that it is made by tailor A = 200/800 = 1/4 = 0.25
Similarly, P(z/Z2) = 300/800 = 3/8 = 0.375 and P(z/Z3) = 0.375
Required probability = Probability that the shirt is made by tailor C given that the shirt is defective = P(Z3/Z)



Required probability = 600/1175 = 24/47
{Baye's Theorem is given by P(Ei/A) 

Test Level 3: Probability - 2 - Question 6

Directions: Study the following information and answer the given question.

Three friends A, B and C do air rifle shooting as a hobby and compete against each other in friendly competitions. On a Sunday morning, they go to an empty tract of land along with their air rifles to shoot at clay practice targets from the distances of 100 m and 200 m. A hits his target 95% of the times from a distance of 100 m and 70% times from a distance of 200 m. The same figures for B are 85% and 80%, respectively. The same figures for C are 90% and 75%, respectively. They put in numbered chits (1, 2 and 3) to be drawn to decide the order of shooting and then each of them shoots at the targets turn-wise according to the chit number, first from 100 m and then from 200 m. They decide on a point system with 1 point for hitting the target from a distance of 100 m and 2 points for hitting it from a distance of 200 m with the winner getting a round of drinks paid by the other two.

Q. What is the probability that the order of shooting is ABC?

Detailed Solution for Test Level 3: Probability - 2 - Question 6

Total number of possible ways = 3!
Number of favourable cases = 1
Thus, required probability = 1/3! = 1/6

Test Level 3: Probability - 2 - Question 7

Directions: Study the following information and answer the given question.

Three friends A, B and C do air rifle shooting as a hobby and compete against each other in friendly competitions. On a Sunday morning, they go to an empty tract of land along with their air rifles to shoot at clay practice targets from the distances of 100 m and 200 m. A hits his target 95% of the times from a distance of 100 m and 70% times from a distance of 200 m. The same figures for B are 85% and 80%, respectively. The same figures for C are 90% and 75%, respectively. They put in numbered chits (1, 2 and 3) to be drawn to decide the order of shooting and then each of them shoots at the targets turn-wise according to the chit number, first from 100 m and then from 200 m. They decide on a point system with 1 point for hitting the target from a distance of 100 m and 2 points for hitting it from a distance of 200 m with the winner getting a round of drinks paid by the other two.

Q. What is the probability that all three of them will end up shooting 1 target only, with all three of them shooting the target of the same points weightage?

Detailed Solution for Test Level 3: Probability - 2 - Question 7

One possibility is that all three of them hit the 1st target and miss out on the 2nd target.
The probability for this is 0.95 × 0.30 × 0.85 × 0.20 × 0.90 × 0.25 = 0.01090125
The 2nd possibility is that all of them hit only the 2nd target and miss out on the 1st.
The probability for the same is 0.05 × 0.7 × 0.15 × 0.80 × 0.10 × 0.75 = 0.000315
Thus, required probability = 0.01090125 + 0.000315 = 0.01121625 ~ 0.011

Test Level 3: Probability - 2 - Question 8

A bag X contains 3 white and 2 black balls. Another bag Y contains 2 white and 4 black balls. A bag and a ball out of it are picked at random. Find the probability that the ball is of white colour.

Detailed Solution for Test Level 3: Probability - 2 - Question 8

A white ball can be drawn in two mutually exclusive ways.
(i) Selecting bag X and then drawing a white ball from it.
(ii) Selecting bag Y and then drawing a white ball from it.
Let E1: event of selecting bag X
E2: event of selecting bag Y
A: event of drawing white ball
∴ P(E1) = P(E2) = 1/2
Now, P(A/E1) = Probability of drawing a white ball when bag X is chosen = 3/5
P(A/E2) = Probability of drawing a white ball when bag Y is chosen = 2/6
P(white ball) = P(A) = P(E1) P(A/E1) + P(E2) P(A/E2) (Law of total probability)

Test Level 3: Probability - 2 - Question 9

Directions: Read the given information and answer the following:

Two friends A and B play a game of chance. A person tosses a coin and if the coin shows a heads, he rolls a die. He does this twice and the score is the total of the numbers that the die throws up in both the rolls, if at all. If a person ends up with a tails on tossing the coin, he does not roll the die and his score is entered as 0. The winner of the game is the person whose score is the maximum.

Q. What is the probability that A wins the game with a score of 2 and B ends up with a score of 0?

Detailed Solution for Test Level 3: Probability - 2 - Question 9

There are 3 favourable cases for the same which shall be discussed one by one.
The 1st case is in which A scores a 1 in both the rolls and B does not get to roll the die.
The probability for the same is (1/2)(1/6)(1/2)(1/6)(1/2)(1/2) = 1/576
The 2nd case is in which A gets a score of 2 in his first roll, gets no score in the 2nd roll and B gets no score in either of his rolls.
The probability for the same is (1/2)(1/6)(1/2)(1/2)(1/2) = 1/96
The 3rd case is in which A gets a score of 0 in the 1st roll, gets a score of 2 in the 2nd roll and B gets no score in either of his rolls.
The probability for the same is (1/2)(1/2)(1/6)(1/2)(1/2) = 1/96
Thus, the required probability is 1/96 + 1/96 + 1/576 = 13/576

Test Level 3: Probability - 2 - Question 10

Directions: Read the given information and answer the following:

Two friends A and B play a game of chance. A person tosses a coin and if the coin shows a heads, he rolls a die. He does this twice and the score is the total of the numbers that the die throws up in both the rolls, if at all. If a person ends up with a tails on tossing the coin, he does not roll the die and his score is entered as 0. The winner of the game is the person whose score is the maximum.

Q. What is the probability that A ends up with an odd score without scoring a 0 in any of his rolls?

Detailed Solution for Test Level 3: Probability - 2 - Question 10

There shall be two cases for the same.
Case I: A scores an even number in his first roll and an odd number in his second roll, the probability of which is (1/2)(3/6)(1/2)(3/6) = 1/16
Case II: A scores an odd number in his first roll and an even number in his second roll, the probability of which is (1/2)(3/6)(1/2)(3/6) = 1/16
Thus, the required probability is 1/16 + 1/16 = 1/8

5 docs|272 tests
Information about Test Level 3: Probability - 2 Page
In this test you can find the Exam questions for Test Level 3: Probability - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Test Level 3: Probability - 2, EduRev gives you an ample number of Online tests for practice

Top Courses for CAT

Download as PDF

Top Courses for CAT