Question

1. A student is taking a multiple-choice exam in which each question has four choices. Assume that the student has no knowledge of the correct answers to any of the questions. She has decided on a strategy
in which she will place four 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 five multiple choice questions on the exam. What is the probability that she will get:

a. Five questions correct?

b. At least four questions correct?

c. No questions correct?

d. No more than two questions correct?

Answer 1

Answer:

a) 0.001 = 0.1% probability that she will get five questions correct.

b) 0.0156 = 1.56% probability that she will get at least four questions correct.

c) 0.2373 = 23.73% probability that she will get no questions correct.

d) 0.8965 = 89.65% probability that she will get no more than two questions correct.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either she gets it correct, or she does not. The probability of getting a question correct is independent of any other question, which means that the binomial probability distribution is used 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.

There are five multiple choice questions on the exam.

This means that $n = 5$

She has decided on a strategy in which she will place four 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.

This means that $p = \frac{1}{4} = 0.25$

a. Five questions correct?

This is $P(X = 5)$. So

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

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

0.001 = 0.1% probability that she will get five questions correct.

b. At least four questions correct?

This is:

$P(X \geq 4) = P(X = 4) + P(X = 5)$

So

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

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0146 + 0.001 = 0.0156$

0.0156 = 1.56% probability that she will get at least four questions correct.

c. No questions correct?

This is P(X = 0). So

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

$P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373$

0.2373 = 23.73% probability that she will get no questions correct.

d. No more than two questions correct?

This is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$. So

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

$P(X = 0) = C_{5,0}.(0.25)^{0}.(0.75)^{5} = 0.2373$

$P(X = 1) = C_{5,0}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2373 + 0.3955 + 0.2637 = 0.8965$

0.8965 = 89.65% probability that she will get no more than two questions correct.