Question

On a multiple-choice test, each question has 4 possible answers. A student does not know

Answer 1

Answer:

42.1875% probability that the student gets all three questions wrong

Step-by-step explanation:

For each question, there are only two possible outcomes. Either the student gets it wrong, or he does not. The probability of the student getting a question wrong is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probabily 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.

On a multiple-choice test, each question has 4 possible answers.

One of these options is correct and the other 3 are wrong. We want to find the probability of getting questions wrong. So $p = \frac{3}{4} = 0.75$

Three question:

This means that $n = 3$

What is the probability that the student gets all three questions wrong?

This is P(X = 3).

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

$P(X = 3) = C_{3,3}.(0.75)^{3}.(0.25)^{0} = 0.421875$

42.1875% probability that the student gets all three questions wrong