Solution: 2019 Fall Final - 11

Author: Michiel Smid

Question

Assume you write a multiple-choice exam that has 25 questions. For each question, four options are given to you, and exactly one of these options is the correct answer. Assume that you answer each question uniformly at random, where the choices for different questions are independent of each other. What is the probability that you have exactly 17 correct answers?

(a)

${25 \choose 17} \cdot \left( 1 \middle/ 4 \right)^8 \cdot \left( 3 \middle/ 4 \right)^{17}$

(b)

${25 \choose 17} \cdot \left( 1 \middle/ 4 \right)^{17}$

(c)

${25 \choose 17} \cdot \left( 3 \middle/ 4 \right)^{17}$

(d)

${25 \choose 17} \cdot \left( 1 \middle/ 4 \right)^{17} \cdot \left( 3 \middle/ 4 \right)^8$

Solution

Let S be the set of all possibilities
$ |S| = 4^{25} $
Let A be the set where we answer 17 correctly
We choose 17 of the 25 questions to be correct: $ \binom{25}{17} $
The remaining 8 each have 3 possible incorrect answers: $ 3^{8} $
$ |A| = \binom{25}{17} \cdot 3^{8} $

$ Pr(A) = \frac{|A|}{|S|} $

$ Pr(A) = \frac{\binom{25}{17} \cdot 3^{8}}{4^{25}} $

$ Pr(A) = \binom{25}{17} \cdot {( \frac{3}{4})}^8 \cdot {( \frac{1}{4})}^{17} $