Solution: 2018 Fall Final - 23

Author: Michiel Smid

Question

The final exam for COMP 2804 has 25 multiple-choice questions. For each question, there are 4 possible answers, exactly one of which is correct. Michiel chooses a positive real number $z$ and uses the following marking scheme: For each correct answer, a student receives 1 mark, whereas for each incorrect answer, the student receives $-z$ marks.
Jim is one of the students and answers the 25 questions, by choosing a uniformly random answer for each question; the choices are independent of each other.
Define the random variable

X = the number of marks that Jim recevies.

For what value of $z$ is the expected value $\mathbb{E}(X)$ equal to 0?
Hint: Use the Linearity of Expectation.

(a)

$z = 3/4$

(b)

$z = 1/3$

(c)

$z = 1/4$

(d)

$z = 1/2$

Solution

Alright. So let’s just linear algebra the hell outta this question

Let $X_i$ be 1 if the student gets the question right and -z if the student gets the question wrong

$X_i = 1$ if the student gets the question right
$ Pr(X_i = 1) = \frac{1}{4} $
$X_i = -z$ if the student gets the question wrong
$ Pr(X_i = -z) = \frac{3}{4} $

$ 0 = \sum_{k=1}^{25} ( 1 \cdot Pr(X_i = 1) ) + \sum_{k=1}^{25} ( -z \cdot Pr(X_i = -z) ) $

$ 0 = \sum_{k=1}^{25} ( 1 \cdot \frac{1}{4} ) + \sum_{k=1}^{25} ( -z \cdot \frac{3}{4} ) $

$ 0 = 25 \cdot ( 1 \cdot \frac{1}{4} ) + 25(-z \cdot \frac{3}{4}) $

$ 0 = 25 \cdot \frac{1}{4} - 25z \cdot \frac{3}{4} $

$ 0 = \frac{25}{4} - \frac{75z}{4} $

$ 0 = 25 - 75z $

$ 75z = 25 $

$ z = \frac{25}{75} $

$ z = \frac{1}{3} $