[BruceZ]

[ QUOTE ]
There are four possibilities. AB, A'B, AB' and A'B'. You have four equations:

AB+AB' = 0.35
AB+A'B = 0.70
A'B' = 0.2
AB+A'B+AB'+A'B' = 1

If we add the first three and subtract the fourth, we get:

AB = 0.35 + 0.70 + 0.20 - 1 = 0.25.

To check, that means:

AB = 0.25
AB' = 0.10
A'B = 0.45
A'B' = 0.20

Those add up to 1 and meet the stated conditions.

[/ QUOTE ]

I think you read his problem as P(A) = .5*P(B) = .35, which would make P(A) = 0.35 and P(B) = 0.7. If you look at the spacing in the original text box (which you get when you quote it) it says P(A) = 0.5 and P(B) = 0.35.

In either case, we can easily solve this without setting up simultaneous equations. We know that P(A) + P(B) computes the probability of the union of A and B plus the intersection of A and B since the intersection is double counted. In other words:

P(A or B) = P(A) + P(B) - P(A and B)

P(A) + P(B) = P(A or B) + P(A and B)

so

P(A) + P(B) + P(A' and B') = P(A or B) + P(A and B) + P(A' and B')

P(A) + P(B) + P(A' and B') = 1 + P(A and B)

Since P(A' and B') = 0.2, if P(A) = 0.5 and P(B) = 0.35 then this equals 1.05, and P(A and B) = 0.05. However, if P(A) = 0.35 and P(B) = 0.7, then this equals 1.25, and P(A and B) = 0.25 as Aaron got.