# GMAT Probability Translation

by on September 2nd, 2012

Translating word problems into algebra is a staple skill of GMAT test-takers, one that underlies countless problems in practice and on Test Day. But some challenging translations occur as part of probability and combinatorics problems. That’s because a pair of the most basic words in the English language, “And” and “Or,” suddenly become overburdened with mathematical significance.

“And” is the simpler of the two. When “And” represents independent choices—cases in which one option or arrangement has no impact on the other choice—just multiply the outcomes. For instance:

“The number of ways to purchase three board games and two video games” is an independent choice. The board games we pick have no impact on the video games we pick. So, to translate: [The number of ways to purchase three board games] × [the number of ways to select two video games]. Of course, we’d need the combination formula to find actual values—but we’d know what to do with those values once we got them.

“Or” is a little more complicated. It’s confusing even in conversation, after all—if I say that you can have cake or ice cream for dessert, can you have both if you want? When you CAN have both, you can treat the problem similarly to an overlapping sets problem. But in most cases on the GMAT, the “Or”s will be mutually exclusive—for instance, if you want to know the odds of drawing a heart or a diamond out of a deck of cards, there is no card that is both a heart or diamond.

A mutually exclusive OR can be translated as a “plus.” That’s all you have to do. So:  “The probability of drawing a heart or a diamond from a deck of cards,” which is the odds of one of two mutually exclusive events occurring, translates to: [The probability of drawing a heart] + [The probability of drawing a diamond].

Today’s problem of the day hinges on those same ideas. Read carefully—you’re solving for the odds of one of two outcomes (an OR), but each of those two outcomes is the specific result of two independent events (an AND). Be systematic in your translation, and I’m sure you’ll get the right result.

## 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:

Step 1: Analyze the Question

This is a complex question, 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.

Step 2: State the Task

Determine the probabilities of the two scenarios above and add them together.

Step 3: Approach Strategically

Let’s start with (i) and find the probability that a senior is chosen from Room A and a nonsenior is chosen from Room B. 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 nonsenior is chosen from Room A and a senior is chosen from Room B.

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