Question

There are n questions on a multiple choice exam, and for each question, there are four choices. To pass the exam, one must correctly answer at least 70% of the questions. The student has not studied, so he/she has to resort to guessing on every question.
a. Find the probability of the student passing for n = 10.b. Find the expected number of questions answered correctly for n = 20.c. Find the variance for the number of questions answered correctly for n = 10.

Answer 1

Answer:

a) 0.352% probability of the student passing

b) 5

c) 1.875

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student answers it correctly, or he answers it wrong. The probability of correctly answering a question is independent from the probability of correctly answering other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The variance of the binomial distribution is:

$V(X) = np(1-p)$

For each question, there are four choices.

One choice is correct, so $p = \frac{1}{4} = 0.25$

a. Find the probability of the student passing for n = 10.

0.7*10 = 7

This is $P(X \geq 7)$ when $n = 10$

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.25)^{7}.(0.75)^{3} = 0.0031$

$P(X = 8) = C_{10,8}.(0.25)^{8}.(0.75)^{2} = 0.0004$

$P(X = 9) = C_{10,9}.(0.25)^{9}.(0.75)^{1} = 0.00002$

$P(X = 10) = C_{10,10}.(0.25)^{10}.(0.75)^{0} \cong 0$

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0031 + 0.0004 + 0.00002 = 0.00352$

0.352% probability of the student passing

b. Find the expected number of questions answered correctly for n = 20.

Here we have n = 20. So

$E(X) = np = 20*0.25 = 5$

c. Find the variance for the number of questions answered correctly for n = 10.

Here we have n = 10.

$V(X) = np(1-p) = 10*0.25*0.75 = 1.875$