If 5 integers are to be selected at random GMAT Probability Question

If 5 integers are to be selected at random and without replacement from the list consisting of the 20 integers from 1 to 20, inclusive, what is the probability that 2 of the 5 selected integers will be 10 and 20 ?

A. 2/5

B. 1/4

C. 1/10

D. 1/19

E. 1/60

OK, I’ll be honest. This is a bit of a loopy question. The writers of the GMAT have been trying for many years to figure out a way to combine Probability and Combinatorics questions, yet with minimal success.

(And then I need to type “if 5 integers are to be selected at random” so the overlords will think this article is relevant.)

As I would normally say, if you’re trying to combine Probability and Combinatorics, you’re making a mistake. The only exception to this would be to apply the Mississippi rule when counting possibilities for the number of ways you can make a selection.

Ultimately, that’s what this question is doing, but the testwriters have done a pretty remarkable job of messing with one’s head in the process. It’s a simple solution, but arriving at the method isn’t immediately obvious.

Let’s dive in.

How to Use Mississippi Rule

If you need a quick refresher, go to this article.

What we need to do here is to think about the different numbers as selections with particular restrictions. Let’s say that picking the number 10 is “A” and picking the number 20 is “B.” These are specific, and the others have no restrictions.

This means that we have a possible order of, for example, ABCCC.

However, we could select any number of ways, so the actual number of orders here, by Mississippi Rule, is

Note that the 1! terms are not strictly necessary, but I’m putting them here to mark A and B given that each only appears one time.

This gives us the number of orders (distributions). And then I type the if 5 integers are to be selected at random business again…

Number of Distributions x Probability per Distribution

Remember that in cases like this, the total Probability is calculated by the number of orders (distributions) multiplied by the Probability of each distribution. Read more about the process here.

Now we simply need to figure out the Probability for each distribution. All we need to do is take one sample case because the distributions will all have the same probability.

The sample case would be this. Note that in the first two cases, we MUST select the given number so the top line of the fraction is 1. Of course if we select these values first, then they will be removed from the pool and we have a 100% Probability for the three C-terms:

…is therefore our Probability per Distribution.

If we multiply the number of Distributions (20) by the Probability per Distribution (\frac{1}{19}\frac{1}{20}), we find:

The answer is (D).