# Robert is writing a test with 20 multiple choice questions, each with 4 possible answers. If he selects his answers by random guessing, how many questions is he expected to answer correctly?

Nov 7, 2017

5

#### Explanation:

For this problem, the following conditions are all true:

• There are a fixed number of "trials" (20 questions to be precise)
• The probability of getting a question correct is the same for every question (since all questions have 4 answers with only 1 being correct)
• Presumably all questions are independent of each other (since we were not told otherwise in the problem statement)

Since all of these conditions hold true, we can consider this to be a binomial random variable problem.

Let $p$ be the probability that Robert gets a single question correct on the exam. Since there are 4 possible answers to a single question, and only 1 of them is correct, then by pure random guessing we see that $p = \frac{1}{4}$.

Let $X$ be the number of questions answered correctly by guessing. Then $X \text{ ~ BIN} \left(n = 20 , p = \frac{1}{4}\right) .$

With $n$ independent questions on the exam, the expected number of correct answers from pure random guessing is given by:

$\text{E} \left[X\right] = n \cdot p$

In this case, since $p = \frac{1}{4}$ and $n = 20$:

$\text{E} \left[X\right] = 20 \cdot \left(\frac{1}{4}\right) = 5$

Robert can expect to get 5 questions correct by random guessing.