Question

A quiz consists of six multiple choice questions. Each question has four choices. A student who forgot to study guesses randomly on every question. What is the probability that the student answers at most four questions correctly?

Answer 1

Answer:

0.9945 = 99.45% probability that the student answers at most four questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student answers correctly, or he/she does not. The probability of the student answering a question 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.

A quiz consists of six multiple choice questions.

This means that $n = 6$

Each question has four choices. Student guesses randomly.

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

What is the probability that the student answers at most four questions correctly?

This is:

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$. So

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

$P(X = 5) = C_{6,5}.(0.25)^{5}.(0.75)^{1} = 0.0053$

$P(X = 6) = C_{6,6}.(0.25)^{6}.(0.75)^{0} = 0.0002$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.0053 + 0.0002 = 0.0055$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.0055 = 0.9945$

0.9945 = 99.45% probability that the student answers at most four questions correctly