GATE Mock Test

Last Updated: 17th October, 2023

1. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

a. 564

b. 645

c. 735

d. 756

Solution:

d. 756

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

2. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

a. 210

b. 1050

c. 25200

d. 21400

e. None of these

Solution:

c. 25200

3. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

1. 810
2. 1440
3. 2880
4. 50400
5. 5760

Solution:

d. 50400

4. From the letters of the word ′DAUGHTER′, how many words can be formed each consisting of 2 vowels and 3 consonants?

1. 32
2. 15
3. 30
4. 16
5. None of these

Solution:

c. 30

DAUGHTER

A,U,E are vowels

D,G,H,T,R are consonants

We want to find words consisting of 2 vowels 3 consonants

Number of words possible =

The result of (3C2 \times 5C3) is equal to 30.

5. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

1. 159
2. 194
3. 205
4. 209
5. None of these

Solution:

d. 209

6. Box A contains 2 white and 3 red balls and box B contains 4 white and 5 red balls. One ball is drawn at random from one of the boxes and is found to be red. Then, the probability that it was from box B, is

1. 25/52
2. 21/52
3. 7/52
4. 15/52
5. None of the above/More than one of the above.

Solution:

a.25/52

Box A contains 2 white and 3 red balls, and Box B contains 4 white and 5 red balls.

Let's define the events:

A: The ball is drawn from Box A.

B: The ball is drawn from Box B.

R: The ball drawn is red.

Given:

Substituting these values into the formula:

Simplify and calculate to find the probability that the red ball was drawn from Box B.

7. A box ′A′ contains 2 white, 3 red and 2 black balls. Another box ′B′ contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box ′B′ is

1. 7/16
2. 9/32
3. 7/8
4. 9/16

Solution:

a. 7/16

We want to find the probability that both balls are drawn from box B, given that one ball is white and the other is red.

Let's define the following events:

• A: The event of selecting box A.
• B: The event of selecting box B.
• W: The event of drawing a white ball.
• R: The event of drawing a red ball.

We want to find  , the probability of drawing both balls from box B, given that we have drawn one white and one red ball.

First, let's find the overall probability of drawing one white and one red ball regardless of the box:

P(A) is the probability of selecting box A, which is 1/2.

is the probability of drawing one white and one red ball from box A, which is

P(B) is the probability of selecting box B, which is also 1/2.

is the probability of drawing one white and one red ball from box B, which is

Now, calculate :

Now, let's calculate using Bayes' theorem:

Substitute the values:

So, the correct probability that both balls are drawn from box B, given that one ball is white and the other is red, is

7. In a dataset of exam scores, the mean score is 75, and the standard deviation is 10. If the data follows a normal distribution, what percentage of students scored above 85?

a. 10% 

b. 16% 

c. 32% 

d. 50%

Solution

b. 16%

Explanation To find the percentage of students who scored above 85 in a normal distribution, you can use z-scores. First, calculate the z-score for 85:

_z_=(85−75)/10=1

Then, find the area to the right of _z_=1 in the standard normal distribution table.

The area to the right of z = 1 is 0.8413. Now subtract 1 - 0.8413 to get the value for P(X>85) which is 0.1587.

This area corresponds to the percentage of students who scored above 85, which is approximately 16%.

9. In a medical test for a rare disease, the probability of a false positive is 5%, and the probability of a false negative is 2%. If the disease occurs in 1% of the population, what is the probability that a person who tests positive actually has the disease?

a. 98% b. 17% c. 30% d. 5%

Solution

b. 17%

Explanation :

To find the probability that a person who tests positive actually has the disease, you can use Bayes' theorem. Let's define the following events:

A: The person has the disease (Disease is present). B: The person tests positive.

We want to find P(A | B), the probability that the person has the disease given that they test positive.

Bayes' theorem states:

In this case:

is the probability of testing positive given that the person has the disease. This is the true positive rate, which is (1 - {false negative rate}). So,  

P(A) is the probability that a person has the disease, which is 1% or 0.01.

P(B) is the probability of testing positive, which can be calculated using the law of total probability: 

Here:

• is the probability of testing positive given that the person does not have the disease. This is the false positive rate, which is 5% or 0.05.
• is the probability that a person does not have the disease, which is the complement of

Now, plug these values into Bayes' theorem:

Calculate the numerator and denominator:

Now, calculate the denominator:

Finally, calculate \(P(A | B)\):

So, the probability that a person who tests positive actually has the disease is approximately 17%.

10. In a survey of a population, the average income is $50,000, and the standard deviation is $8,000. If we randomly select 25 individuals from this population, what is the probability that their average income is less than $48,000?

a. 0.1056 b. 0.8413 c. 0.0228 d. 0.9772

Solution

Answer a. 0.1056

Explanation: To find the probability that the average income of 25 individuals is less than $48,000, you can use the Central Limit Theorem. First, find the z-score:

Next, find the cumulative probability to the left of _z_=−1.25 in the standard normal distribution table. The probability is approximately 0.1056.