Math Help - P(A|B)

I hate Statistics.

Anyway.

Two independant events are such that P(A)=0.2, P(B)=0.25, P(AuB)=0.4, P(AnB)=0.05

All I have to figure out is:

Given that exactly one event has occured, find the probablity that A occurs.

I know the formula for given is P(X|Y)=P(XnY)/P(Y) and the answer is 0.2*0.75/0.35

I just want to know where each of the numbers come from. Mainly the 0.75. Thanks.

Originally Posted by StephenPoco

I hate Statistics.

Anyway.

Two independant events are such that P(A)=0.2, P(B)=0.25, P(AuB)=0.4, P(AnB)=0.05

All I have to figure out is:

Given that exactly one event has occured, find the probablity that A occurs.

I know the formula for given is P(X|Y)=P(XnY)/P(Y) and the answer is 0.2*0.75/0.35

I just want to know where each of the numbers come from. Mainly the 0.75. Thanks.

Honestly I don't know where the 0.75 comes from, but I get the same answer.

We are looking for the P(only A occurred | (only A occurred) or (only B occurred)).

In other words we want P(A and not B | (A and not B) or (B and not A)).

In other words, P(A and not B | (A or B) - (A and B)).

With Venn diagrams it looks really simple.

So we get (0.2-0.05) / (0.4-0.05) = 0.15/0.35 = 3/7.

Yeah, I know that it's 0.15/0.35. It's just I can't seem to figure out where the 0.75 came from which really threw me off track.

Originally Posted by StephenPoco

Yeah, I know that it's 0.15/0.35. It's just I can't seem to figure out where the 0.75 came from which really threw me off track.

Well P(A and not B) is 75% of P(A).

But why anyone would try to solve it that way is beyond me.