Question

Jane must get at least three of the four problems on the exam correct to get an A. She has been able to do 80% of the problems on old exams, so she assumes that the probability she gets any problem correct is 0.8. She also assumes that the results on different problems are independent. (a) What is the probability she gets an A? (b) If she gets the first problem correct, what is the probability she gets an A?

Answer 1

a) There is n 81.92% probability that she gets an A.

b) If she gets the first problem correct, there is an 89.6% probability that she gets an A.

Explanation:

For each question, there are only two possible outcomes. Either the answer is correct, or it is not. This means that we can solve this problem using binomial distribution probability concepts.

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).\pi^(x).(1-\pi)^(n-x)`

In which
`C_(n,x)` is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And
`\pi` is the probability of X happening.

For this problem, we have that:

The probability she gets any problem correct is 0.8, so
`\pi = 0.8`.

(a) What is the probability she gets an A?

There are four problems, so
`n = 4`

Jane must get at least three of the four problems on the exam correct to get an A.

So, we need to find
`P(X \geq 3)`

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

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

`P(X = 3) = C_(4,3).(0.80)^(3).(0.2)^(1) = 0.4096`

`P(X = 4) = C_(4,4).(0.80)^(4).(0.2)^(0) = 0.4096`

`P(X \geq 3) = P(X = 3) + P(X = 4) = 2*0.4096 = 0.8192`

There is n 81.92% probability that she gets an A.

(b) If she gets the first problem correct, what is the probability she gets an A?

Now, there are only 3 problems left, so
`n = 3`

To get an A, she must get at least 2 of them right, since one(the first one) she has already got it correct.

So, we need to find
`P(X \geq 2)`

`P(X \geq 3) = P(X = 2) + P(X = 3)`

`P(X = 2) = C_(3,2).(0.80)^(2).(0.2)^(1) = 0.384`

`P(X = 4) = C_(3,3).(0.80)^(3).(0.2)^(0) = 0.512`

`P(X \geq 3) = P(X = 2) + P(X = 3) = 0.384 + 0.512 = 0.896`

If she gets the first problem correct, there is an 89.6% probability that she gets an A.