Question

Assume 75% of the AP stats students studied for this test. If 40% of those who studied get an A but only 10% of those who did not study get A, what is the probability that someone who gets an A actually studied for the test?

Answer 1

Let `A=` the student get an A in test
Let `B=` the student studied for test

We know that `P(B)=0.75` and `P(A//B)=0.40`

We know also that `P(A//B)=(P(AnnB))/(P(B))=0.40`

We are looking for `P(AnnB)=0.40xx0.75=0.3`

Answer 2

The probability is (approximately) 92.31%.

Explanation:

Let `A` be the event that the student got an A on the test.
Let `B` be the event that the student studied.

It sounds like we are seeking the conditional probability of, "studied for the test GIVEN THAT they got an A".

This is written as `"P"(B | A)`, and it is read as "probability of B given A."

We are told:

`"P"(B) = 0.75` which means `"P"(B^"C") = 0.25`
`"P"(A|B)=0.4`
`"P"(A|B^"C") = 0.1`
(the `""^"C"` means "complement")

and we want to find `"P"(B|A)`. By definition,

`"P"(B|A) = ("P"(B nn A))/("P"(A))`

Using Bayes' Theorem, we can rewrite this as

`"P"(B|A) = ("P"(A|B)"P"(B))/("P"(A|B) "P"(B)+"P"(A|B^"C") "P"(B^"C"))`

Every probability on the right side is now something we know. We plug these values in:

`"P"(B|A) = (0.4 xx 0.75)/(0.4 xx 0.75"  " +"  "0.1 xx 0.25)`

`color(white)("P"(B|A)) = (0.3)/(0.3+0.025)`

`color(white)("P"(B|A)) = (0.3)/(0.325)`

`color(white)("P"(B|A)) = 12/13" " ~~ 0.9231`

So there is a 92.31% chance that, given that a student got an A, they also studied.