# A student is taking a multiple-choice exam in which each question has five choices. Assuming that...

## Question:

A student is taking a multiple-choice exam in which each question has five choices. Assuming that she has no knowledge of the correct answers to any of the questions, she has decided on a strategy in which she will place five balls (marked A, B, C, D, and E) into a box. She randomly selects one ball for each question and replaces the ball in the box. The marking on the ball will determine her answer to the question. There are six multiple-choice questions on the exam. Complete parts (a) through (d) below.

A. What is the probability that she will get six questions correct? (Round to four decimal places as needed.)

B. What is the probability that she will get at least five questions correct? (Round to four decimal places as needed.)

C. What is the probability that she will get no questions correct? (Round to four decimal places as needed.)

D. What is the probability that she will get no more than two questions correct? (Round to four decimal places as needed.)

## The Binomial Probability Distribution

Since there are n multiple choice questions, and a probability, p, of getting the newer correct, then the following problem clearly follows the binomial distribution with parameter n and p. Here, we can use the following functions:

- {eq}P(X=x) = \binom{n}{x} (p)^x (1-p)^{n-x} {/eq} and,

- {eq}P(a \lt x \lt b) = \sum_{x=a}^{b}{ \binom{n}{x} (p)^x (1-p)^{n-x}} {/eq}

## Answer and Explanation:

Here we have that n = 6 and the probability of getting an answer correct is p = 0.20. Thus we can answer the following

A) The probability of getting exactly 6 correct is given as

{eq}P(X=6) = \binom{6}{6} (0.20)^6 (0.80)^{0} = 0.000064 {/eq}

B) The probability of getting at least 5 is given as

{eq}P(X \geq 5) = P(5 \lt x \lt 6) = \sum_{x=5}^{6}{ \binom{6}{x} (0.2)^x (0.8)^{6-x}} = 0.0016 {/eq}

C) The probability of getting none correct is given as:

{eq}P(X=0) = \binom{6}{0} (0.2)^0 (0.8)^{6} = 0.2621 {/eq}

D) The probability that she will get no more than 2 questions is given as

{eq}P(X \leq 2) = P(0 \lt x \lt 2) = \sum_{x=0}^{2}{ \binom{6}{x} (0.2)^x (0.8)^{6-x}} = 0.9011 {/eq}

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