Question

Suppose a student takes a multiple-choice exam composed of 8 questions with 5 possible answers to each question, exactly one of which is correct. The student has no clue as to the correct answer to any of the question! and merely guesses. [You can use binompdf or binomcdf for a), b), and c).] {5.4)

a) What is the probability that the student will get exactly 6 answers correct, which is necessary for a grade of 751 (5 points)
b) What is the probability of the student getting 3 or less answers correct? [5 points)
c) What is the probability of the student getting 4 or more answers correct? [Hint: the complement of this event is getting 3 or less answers correct.] [4 point

Answer 1

a) 0.11% probability that the student will get exactly 6 answers correct

b) 94.37% probability of the student getting 3 or less answers correct

c) 5.63% probability of the student getting 4 or more answers correct

Explanation:

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.

Suppose a student takes a multiple-choice exam composed of 8 questions with 5 possible answers to each question, exactly one of which is correct.

This means that
`n = 8, p = (1)/(5) = 0.2`

a) What is the probability that the student will get exactly 6 answers correct, which is necessary for a grade of 751 (5 points)

This is P(X = 6).

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

`P(X = 6) = C_(8,6).(0.2)^(6).(0.8)^(2) = 0.0011`

0.11% probability that the student will get exactly 6 answers correct

b) What is the probability of the student getting 3 or less answers correct? [5 points)

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

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

`P(X = 0) = C_(8,0).(0.2)^(0).(0.8)^(8) = 0.1678`

`P(X = 1) = C_(8,1).(0.2)^(1).(0.8)^(7) = 0.3355`

`P(X = 2) = C_(8,2).(0.2)^(2).(0.8)^(6) = 0.2936`

`P(X = 3) = C_(8,3).(0.2)^(3).(0.8)^(5) = 0.1468`

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1678 + 0.3355 + 0.2936 + 0.1468 = 0.9437`

94.37% probability of the student getting 3 or less answers correct

c) What is the probability of the student getting 4 or more answers correct? [Hint: the complement of this event is getting 3 or less answers

`P(X \leq 3) + P(X \geq 4) = 1`

We want
`P(X \geq 4)`

So

`P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.9437 = 0.0563`

5.63% probability of the student getting 4 or more answers correct