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Sample GMAT Question: Probability + Combinations

Written by Sameer Kamat

In another post we covered tips and techniques to manage GMAT quantitative (Maths) questions that involve concepts of Combinations and Probability. In this post on MBA Crystal Ball, the GoGMAT tutors share a sample GMAT quant question that builds up on the theory.


Sample GMAT Maths Question: Combinations and Probability

Although GMAT questions are generally designed to test specific skills, at higher levels, they begin to combine different math concepts within one question. In our previous post we saw how GMAT may combine probability and combinations/permutations in one question; in this post we will analyze a typical GMAT question on this topic.

Here is an example question that you should attempt on your own before following to the answer breakdown below.

A box holds 4 yellow shirts and 4 blue shirts. What is the probability of getting exactly three yellow shirts or exactly three blue shirts when taking out 4 shirts randomly out of the box and returning each shirt before taking out the next one?

A) 1/8

B) 1/4

C) 1/2

D) 2/3

E) 3/4

 

Here’s the explanation:

Before we start, lets break the question down to take out the key information and plan how we will arrive at our answer.

The important parts are:

  • We have 4 yellow shirts (We will label these Y)
  • We have 4 blue shirts (We will label these B)
  • Of all 8 shirts, only 4 are picked
  • Each shirt is returned, so the chance of being picked always remains the same
  • We need to work out exactly 3 yellow shirts OR exactly 3 blue shirts
Ultimately we need to find a probability, so the result will be found using this equation:

Probability of A happening = (number of times A can occur) / ( number of time any outcome can occur)

 

For our specific question this becomes:

Probability of 3Y or 3B = (number of times 3Y or 3B can occur) / ( number of time any outcome can occur)

 

It is clear that we will be looking at either permutations or combinations to calculate the number of times 3Y or 3B can occur (that simply implies the number of time three yellow shirts can occur or the number of time three blue shirts can occur).

Let’s start with the denominator: number of times any outcome can occur.

 

Step 1:

There is a slight complication in this question. We have eight objects, but only four will ever be selected from the box. So, we have to think logically about this.

If we think about it, there are only two possible options for the first selection; it’s either yellow or blue. That gives us two options for the first pick.

The second selection can also have only two equal choices; again either yellow or blue.

For the first selection we have only 2 different outcomes possible. However, because we have 2 different possibilities for the second selection too, we now have four total different possibilities.

The combinations possible could be:

B, Y

B, B

Y, B

Y, Y

 

When we multiply the number of possible outcomes from selection 1, by the number of possible outcomes from selection 2, we get the total number of possible outcomes from steps 1&2.

So for 4 selections, the number of possible outcomes will be:

Total outcomes = 2 × 2 × 2 × 2 = 16

 

So, there are 16 possible outcomes, and we can now update our probability equation to:

Probability of 3Y or 3B = (number of times 3Y or 3B can occur) /16

 

Step 2:

Now we are left with calculating the number of times 3Y or 3B can occur. The way to do this is to split the yellow and blue combinations and work them out separately.

Let’s start with the yellow shirts (Y). We have to calculate how many ways there are to pick 3Y from a selection of 4 shirts. As we are looking at the number of different ways to get the same result (the order of our yellow shirts does not matter), we know we are using the combinations equation:

Combinations = n! / (r! (n-r)!)

n, represents the number of items in the set, so in our case it is 4. r, represents the size of the selection, which in this question is 3.

 

Substituting these values gives:

Combinations = 4!/(3! (4-3)!) = 4!/(3! 1!) = (4×3×2×1) / (3×2×1) = 4

 

So there are four ways to pick exactly three yellow shirts from a selection of four shirts.

As the number of blue shirts is the same, and we have to select the same number of blue shirts (exactly three), it would make sense that there will be 4 different ways to pick exactly three blue shirts too.

That gives us a total of 8 different combinations that give us either exactly three blue shirts or exactly three yellow shirts

 

Going back to our probability equation:

Probability of 3Y or 3B = (number of times 3Y or 3B can occur) / 16

We can now fill in the top part of the fraction, to get:

Probability of 3Y or 3B = 8 /16 = 1 / 2

The answer is therefore: 1/2

 

Summary

This question is tricky as there are a number of steps that you must figure out quickly. The key is really to determine what exactly you have to solve, and then to begin finding the missing bits of data to complete the equation. As always, practice make perfect!


GMAT Preparation Tips by GoGMAT
GoGMAT, founded in 2009, is an adaptive GMAT preparation platform developed by the best instructors in the industry (with 740+ GMAT scores and strong teaching experience). Find out more here: GMAT preparation courses


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Sameer Kamat

About Sameer Kamat

Founder of MBA Crystal Ball | Author of 'Beyond The MBA Hype' | Cambridge MBA Connect with me on Google+ | Twitter @mba_cb


1 Comment

  1. Riffat   |  Sunday, 31 March 2013 at 5:29 pm

    very good example

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