Two events A and B are such that P(A∪B)=0.7 and P(A|B')=0.6. Find P(B).
2026-03-30 02:57:37.1774839457
IB HL Math, Conditional probability question
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According to Bayes' Theorem, we know that $P(A|B') = \frac{P(A\cap B')}{P(B')} = \frac{P(A\cap B')}{1 - P(B)}$. By looking at the Venn diagram we can also see that $P(A\cup B) = P(A\cap B') + P(B)$, which means $P(A\cap B') = 0.7 - P(B)$. We finally have the equation $0.6 = \frac{0.7 - P(B)}{1 - P(B)}$, which we solve to get $P(B) = 0.25$