Solution: 2018 Fall Midterm - 17

Author: Michiel Smid

Question

This midterm has 17 questions. For each question, four options are given. Assume that you answer each question, by choosing one of the four options uniformly at random.
Define the event

A = "you answer exactly 7 questions correctly".

What $\Pr(A)$?

(a)

$\frac{{17 \choose 7}}{4^{17}}$

(b)

$\frac{4^{17}}{{17 \choose 7}}$

(c)

$\frac{{17 \choose 7} \cdot 3^{10}}{4^{17}}$

(d)

$\frac{{17 \choose 7} \cdot 2^{10}}{4^{17}}$

Solution

There are two options to doing this

First, we need to choose 7 questions to answer correctly from the 17 questions: $ \binom{17}{7} $
For each of the 7 questions, there is a $ \frac{1}{4} $ chance of answering correctly: $ {( \frac{1}{4})}^7 $
For each of the 10 questions, there is a $ \frac{3}{4} $ chance of answering incorrectly: $ {( \frac{3}{4})}^{10} $
Pr$(A) = \binom{17}{7} \cdot {( \frac{1}{4})}^7 \cdot {( \frac{3}{4})}^{10} $
Pr$(A) = \binom{17}{7} \cdot {( \frac{1}{4})}^{17} \cdot 3^{10} $
Pr$(A) = \frac{ \binom{17}{7} \cdot 3^{10}}{4^{17}} $
Let A be the event that you answer exactly 7 questions correctly.
Choose 7 questions to answer correctly from the 17 questions: $ \binom{17}{7} $
For each of the 7 questions that you get correct, there is only 1 proper choice: $1 \cdot 1 \cdot ... \cdot 1 = 1^7$
For each of the 10 questions that you get wrong, there are 3 proper choices: $3 \cdot 3 \cdot ... \cdot 3 = 3^{10}$
$|A| = \binom{17}{7} \cdot 1^7 \cdot 3^{10}$
The entire set of all possible combinations is just $|S|=4^{17}$
$ Pr(A) = \frac{|A|}{|S|}$
$ Pr(A) = \frac{ \binom{17}{7} \cdot 1^7 \cdot 3^{10}}{4^{17}} $