All right, listen up. Here’s the deal. We know you guys are out there tearing it up and studying hard for the GMAT, so each week we’re going to break down a tough GMAT problem for you here on the blog. This will be straight to the point, outcomes-focused practice. No fluff. Let’s get down to business.

**Question:**

Each person in Room A is a student, and 1/6 of the students in Room A are seniors. Each person in Room B is a student, and 5/7 of the students in Room B are seniors. If 1 student is chosen at random from Room A and 1 student is chosen at random from Room B, what is the probability that exactly 1 of the students chosen is a senior?

(A) 5/42

(B) 37/84

(C) 9/14

(D) 16/21

(E) 37/42

**Solution:**

For those of you who love probability, you are rejoicing at this question. For everyone else, you are thinking something closer to “Sh@*, I hate these!” Whatever perspective you’re starting with, we’re going to get you over the finish line.

This is a complex question for sure, but it can be broken down into simple steps. As with any probability question, we must first consider all of the scenarios in which the desired outcome can be true. In this question, there are two different ways in which exactly one of two students chosen is a senior. Either (i) a senior is chosen from Room A and a non-senior is chosen from Room B or (ii) a non-senior is chosen from Room A and a senior is chosen from Room B. What we need to do is to determine the probabilities of the two scenarios above and add them together.

Let’s start with (i) and find the probability that a senior is chosen from Room A and a non-senior is chosen from Room B. To do this we’ll find the probability of each and multiply them together (the “and” tells us to multiply here).

The probability that the student chosen from Room A is a senior is 1/6. The probability that the student chosen from Room B is not a senior is 1 – 5/7 = 2/7. So the probability that the student chosen from Room A is a senior and the student chosen from Room B is not a senior is (1/6) x (2/7) = 2/42.

Let’s not simplify this yet, because we can expect that the probability we will find when working with (ii) will also have a denominator of 42.

Now let’s work with (ii). Let’s find the probability that a non-senior is chosen from Room A and a senior is chosen from Room B. To do this we will, again, find the probability of each and multiply them together (the “and” tells us to multiply here).

The probability that the student chosen from Room A is not a senior is 1 – 1/6 = 5/6. The probability that the student chosen from Room B is a senior is 5/7. So the probability that the student chosen from Room A is a not a senior and the student chosen from Room B is a senior is (5/6) x (5/7) = 25/42.

Now we sum the total desired outcomes (the “exactly one” wording in the original question tells us that this is an “or” situation – either this event or that. The “or” tells us to add here).

The probability that exactly one of the students chosen is a senior is (2/42) + (25/42) = 27/42 = 9/14.

**(C) is correct.**