Question

What's the probability that the student will pass the exam by following her strategy?

A student is taking a multiple choice exam in which each question has 4 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 4 balls (marked A, B, C and D) 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 5 multiple choice questions on the exam. Suppose that the exam has 50 multiple choice questions and 30 or more correct answers is a passing score.

Answer 1

roughly `0.0355`

Explanation:

We start with a relation that talks about binomial probability:

`sum_(k=0)^(n)C_(n,k)(p)^k(1-p)^(n-k)=1`

We have `n=50` (there is one sentence in the question that says there are only 5 questions but I'm assuming there was an error and what was meant is 50 questions, which is consistent with the rest of the question).

The probability of getting an answer right by guessing is `p=1/4`.

A passing grade is `k>=30`

The probability of getting a passing score is:

`C_(50,30)(1/4)^30(3/4)^(20)+C_(50,31)(1/4)^31(3/4)^(19)+...+C_(50,50)(1/4)^50(3/4)^(0)`

Instead of summing all this up, I'll use this online calculator tool:

http://stattrek.com/online-calculator/binomial.aspx

We take the box labeled `P(X>=x)`, which shows a value of roughly `0.0355`