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Problem 2

Given the probability of P(A) = .9 and P(B) = .8. Prove that P(AnB) >= .7.

Solution

We know that P(AuB) = P(A) + P(B) - P(AnB) => P(AnB) = P(A) + P(B) - P(AuB).

We also know that P(Set) >= P(AuB).

Therefore P(AnB) = P(A) + P(B) - P(AuB) >= P(A) + P(B) - P(Set)

Which gives us P(A) + P(B) - P(AuB) >= .9 + .8 - 1 = .7

Thus P(AnB) >= .7