Bunuel wrote:

A certain multiple-choice math quiz contains five questions, each with four answer choices: A, B, C, and D. If the correct answer to each question is randomly distributed, which of the following represents the closest approximation of the probability that no two consecutive questions will have the same answer?

A. 25%

B. 32%

C. 40%

D. 48%

E. 75%

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:This question feels like a doozy. Probability questions tend to give fits to even top-performing test-takers. You simply have to remember, however, what probability is:

number of desired outcomes divided by number of possible outcomes

Let's start out by finding the number of total outcomes. There are five questions on the quiz, so let's draw out five slots:

How many possible outcomes are there for the first slot? Well, given that there are four answer choices, A, B, C, and D, there are four possible outcomes.

4_ __ ____ ____ ____

The same will be true for all of the other questions, as each question has exactly four possible results: A, B, C, and D.

4_ _4 4_ 4_ _4

Since there are four possibilities for each of five slots, there are a total of 4 * 4 * 4 * 4 * 4 possibilities, or 4^5 possible outcomes. This number will go in the denominator of our probability fraction.

Now, let's focus on the number of desired outcomes. As it was before, the number of possibilities in the first slot will be four, as the correct answer to the first question could be A, B, C, or D.

4_ ____ ____ ____ ____

Here is where things get a bit tricky. The answer to the second question can be anything except the answer to the first. For example, if the answer to Question 1 is A, then Question 2 can only be B, C, or D.

If the answer to Question 1 is B, then Question 2 can only be A, C, or D. In any case, there are exactly three possibilities for the second slot.

4_ 3_ ____ ____ ____

Based on the answer to Question 2, there are three possibilities for Question 3 (because Question 3 cannot have the same answer as Question 2). Thus, there will also be three possibilities for the fourth and fifth slots, as their only restrictions are not having the same answer as the question immediately preceding it.

4_ 3_ 3_ 3_ 3_

Multiply these together, and you get 4 * 3 * 3 * 3 * 3, or 4 * 3^4 as the number of desired outcomes. So here is the probability fraction, and its subsequent reduction:

desired outcomes = 4 * 3^4

possible outcomes = 4^5

And since you're dividing them, you can factor out the common 4 to get a fraction of (3^4)/(4^4), which is 81/256.

Resist the temptation to do this math and figure out the exact percentage! You know that 80 / 240 = 33.333%, so 81/256 will be slightly less. The only answer that works is 32%, or answer choice B.

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