Are you desperately looking for practice questions on the GMAT club forum? How many questions did you find for the topic you were looking for? Are they any good?

Well, if you are scavenging for good practice questions you’ve just hit a gold mine! Problem solving comprises roughly 50% of the total questions in the GMAT quant section and in this blog, we will show you how to solve **GMAT** probability questions through some quality practice questions and their detailed explanations. Let’s go!

## GMAT Probability Question 1.

A person has 3 children with at least one boy. Find the probability of having at least 2 boys among the children.

**Solution **The event is occurring under the following situations:

- Second is a boy and third is a girl OR
- Second is a girl and third is a boy OR
- Second is a boy and third is a boy

This will be represented by: (1/2) x (1/2) + (1/2) x (1/2) + (1/2) x (1/2) = 3/4

## GMAT Probability Question 2.

Out of 13 applicants for a job, there are 5 women and 8 men. Two persons are to be selected for the job. The probability that at least one of the selected persons will be a woman is:

**Solution****:** The required probability will be given by

First is a woman and Second is a man OR

First is a man and Second is a woman OR

First is a woman and Second is a woman

i.e. (5/13) x (8/12) + (8/13) x (5/12) + (5/13) x (4/12)

= 100/156 = 25/39

Alternatively, we can define the non-event as: There are two men and no women. Then, the probability of the nonevent is

(8/13) x (7/12) = 56/156

Hence, ** P(E) = **(1– 56/156) = 100/156 = 25/39

** [Note: **This is a case of probability calculation where repetition is not allowed.]

## GMAT Probability Question 3.

The probability that ** A **can solve the problem is 2/3 and

**can solve it is 3/4. If both of them attempt the problem, then what is the probability that the problem gets solved.**

*B***Solution **The event is defined as:

** A **solves the problem AND

**does not solve the problem**

*B*OR

A doesn’t solve the problem **AND B **solves the problem

OR

A solves the problem AND *B solves *the problem. Numerically, this is equivalent to:

(2/3) x (1/4) + (1/3) x (3/4) + (2/3) x (3/4)

= (2/12) + (3/12) + (6/12) = 11/12

## GMAT Probability Question 4.

Six positive numbers are taken at random and are multiplied together. Then what is the probability that the product ends in an odd digit other than 5?

**Solution **The event will occur when all the numbers selected are ending in 1, 3, 7, or 9.

If we take numbers between 1 to 10 (both inclusive), we will have a positive occurrence if each of the six numbers selected is either 1, 3, 7, or 9.

The probability of any number selected being either of these 4 is 4/10 (4 positive events out of 10 possibilities) ** [Note: **If we try to take numbers between 1 to 20, we will have a probability of 8/20 = 4/10. Hence, we can extrapolate up to infinity and say that the probability of any number selected ending in 1, 3, 7, or 9 so as to fulfill the requirement is 4/10.

Hence, answer = (0.4)^{6}

Have you taken the GMAT before?

## GMAT Probability Question 5.

The probability that Arjit will solve a problem is 1/5. What is the probability that he solves at least one problem out of ten problems?

**Solution **The non-event is defined as:

He solves no problems i.e. he doesn’t solve the first problem and he doesn’t solve the second problem … and he doesn’t solve the tenth problem.

Probability of non-event = (4/5)^{10}

Hence, the probability of the event is 1-(4/5)”

## GMAT Probability Question 6.

A carton contains 25 bulbs, 8 of which are defective. What is the probability that if a sample of 4 bulbs is chosen, exactly 2 of them will be defective?

**Solution **The probability that exactly two balls are defective and exactly two are not defective will be given by

(4C_{2}) x (8/25) x (7/24) x (17/23) x (16/22)

## GMAT Probability Question 7.

** **Out of 40 consecutive integers, two are chosen at random. Find the probability that their sum is odd.

**Solution **Forty consecutive integers will have 20 odd and 20 even integers. The sum of 2 chosen integers will be odd, only if

- First is even and Second is odd OR
- First is odd and Second is even

Mathematically, the probability will be given by:

**P(First is even) x P(Second is odd) + P(First is odd)** **x P(second is even)**

= (20/40) x (20/39) + (20/40) x (20/39)

= (2 *x *20^{2}/40 x 39) = 20/39

## GMAT Probability Question 8.

An integer is chosen at random from the first 100 integers. What is the probability that this number will not be divisible by 5 or 8?

**Solution: **For a number from 1 to 100 not to be divisible by 5 or 8, we need to remove all the numbers that are divisible by 5 or 8.

Thus, we remove 5, 8, 10, 15, 16, 20, 24, 25, 30, 32 35, 40, 45, 48, 50, 55, 56, 60, 64, 65, 70, 72, 75, 80, 85: 88, 90, 95, 96, and 100.

i.e. 30 numbers from the 100 are removed.

Hence, the answer is 70/100 = 7/10 (required probability)

Alternatively, we could have counted the numbers as the number of numbers divisible by 5 + the number of numbers divisible by 8 – the number of numbers divisible by both 5 or 8.

= 20 + 12 – 2 = 30

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