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 Upper A comma Upper B comma Upper C comma Upper D comma and Upper 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 four ​multiple-choice questions on the exam. Complete parts​ (a) through​ (d) below. a. What is the probability that she will get four questions​ correct? 0.0016 ​(Round to four decimal places as​ needed.) b. What is the probability that she will get at least three questions​ correct? 0.0272 ​(Round to four decimal places as​ needed.) c. What is the probability that she will get no questions​ correct? nothing ​(Round to four decimal places as​ needed.)

Answer 1

n = 4

P = 0.2

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

A) P(X = 4) = 4C4 * (0.2)^4 * (0.8)^0 = 0.0016

B) P(X > 3) = P(X = 3) + P(X = 4)

= 4C3 * (0.2)^3 * (0.8)^1 + 4C4 * (0.2)^4 * (0.8)^0

= 0.0272

C) P(X = 0) = 4C0 * (0.2)^0 * (0.8)^4 = 0.4096