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?

2 Answers
May 21, 2018

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#

May 26, 2018

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.