Question

g Consider a multiple-choice question exam consisting of 20 questions. Assume that each question has five possible choices, and only one of them is correct. If a student is going to guess solutions at the exam, what is the probability that he answers at most three of them are correct

Answer 1

Answer:

0.4114 = 41.14% probability that he answers at most three of them are correct

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent of any other question. This 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.

20 questions:

This means that $n = 20$

Assume that each question has five possible choices, and only one of them is correct.

This means that $p = \frac{1}{5} = 0.2$

What is the probability that he answers at most three of them are correct?

This is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

So

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

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0115 + 0.0576 + 0.1369 + 0.2054 = 0.4114$

0.4114 = 41.14% probability that he answers at most three of them are correct