Question

A quiz has 10 multiple choice questions each with 4 answers that are equally likely to be the correct one. Suppose that the quiz takers need to score 7 corrects out of 10 to pass. Answer the following questions when the quiz taker selects the answers randomly. a) Probability of marking exactly 3 incorrect answers. b) Probability of passing

Answer 1

a) 0.0031 = 0.31% probability of marking exactly 3 incorrect answers.

b) 0.0035 = 0.35% probability of passing.

Explanation:

For each question, there are only two possible outcomes. Either the student chooses the correct answer, or he does not. Questions are independent of each other. So we use the binomial probability distribution 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) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

10 multiple choice questions

This means that
`n = 10`

4 answers that are equally likely to be the correct one. The quiz taker selects the answers randomly.

This means that
`p = (1)/(4) = 0.25`

a) Probability of marking exactly 3 incorrect answers.

3 incorrectly = 7 correctly, so this is P(X = 7).

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

`P(X = 7) = C_(10,7).(0.25)^(7).(0.75)^(3) = 0.0031`

0.0031 = 0.31% probability of marking exactly 3 incorrect answers.

b) Probability of passing

At least 7 correct, so

`P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)`

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

`P(X = 7) = C_(10,7).(0.25)^(7).(0.75)^(3) = 0.0031`

`P(X = 8) = C_(10,8).(0.25)^(8).(0.75)^(2) = 0.0004`

`P(X = 9) = C_(10,9).(0.25)^(9).(0.75)^(1) \approx 0`

`P(X = 10) = C_(10,10).(0.25)^(10).(0.75)^(0) \approx 0`

`P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0031 + 0.0004 + 0 + 0 = 0.0035`

0.0035 = 0.35% probability of passing.