Question

In a multiple choice exam, there are 5 questions and 4 choices for each question (a, b, c, d). Nancy has not studied for the exam at all and decides to randomly guess the answers. What is the probability that: (a) the first question she gets right is the 5th question? (b) she gets all of the questions right? (c) she gets at least one question right?

Answer 1

a) There is a 7.91% probability that the first question she gets right is the 5th question.

b) There is a 0.1% probability she gets all of the questions right.

c) There is a 76.27% probability that she gets at least one question right.

Explanation:

There are only two possible outcomes for each question. Either Nancy gets it right, or she does not. So the binomial probability distribution is used to solve this problem.

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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Each question has 4 choices, one of which is correct. Nancy get's each question. So there is a 1/4 = 0.25 probability she gets it right and a 3/4 = 0.75 probability she gets it wrong.

(a) the first question she gets right is the 5th question?

Nancy has to get the first four question wrong and the 5th right.

For each question, there is a 0.75 probability she gets it wrong and a 0.25 probability she gets it right.

So,

`P = (0.75)^(4)*0.25 = 0.0791`

There is a 7.91% probability that the first question she gets right is the 5th question.

(b) she gets all of the questions right?

This is P(X = 5), with
`n = 5, p = 0.25`.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 5) = C_(5,5).(0.25)^(5).(0.75)^(0) = 0.0001`

There is a 0.1% probability she gets all of the questions right.

(c) she gets at least one question right?

Either she does not get any question right, or she gets at least one question right. The sum of the probabilities of these events is decimal 1. So:

`P(X = 0) + P(X \geq 1) = 1`

`P(X \geq 1) = 1 - P(X = 0)`

In which

`P(X = 0) = C_(5,0).(0.25)^(0).(0.75)^(5) = 0.2373`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2373 = 0.7627`

There is a 76.27% probability that she gets at least one question right.