Problem 2.1 • Find P(A) and P(B) given P(A∩B) and P(A∪B)
With only P(A∩B) and P(A∪B), we can find P(A)+P(B) but not individual values. We need one more constraint (like independence) to solve uniquely.
Any (P(A), P(B)) pair that sums to the required value works.
With independence: P(A∩B) = P(A)·P(B), we get a quadratic!