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Suppose $P(A) = 0.6$ and $P(B) = 0.7$. Find the smallest possible value of $P(A \cap B)$. Instead of deducing this from $P(A \cup B) = P(A)+P(B)-P(A \cap B)$, could we just use the fact that $P(A|B)P(B) = P(A \cap B)$ and $P(B|A)P(A) = P(A \cap B)$? I notice that $(0.6)(0.7)^2 = 0.294$ which is close to $0.3$, the smallest value of $P(A \cap B)$.

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up vote 1 down vote accepted

From the equation in the second line $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ we have $$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$ In order to minimize $P(A \cap B)$, we should maximize $P(A \cup B)$ since both $P(A)$ and $P(B)$ are fixed. The maximum value for $P(A \cup B)$ is $1$, so we have the minimum value of $P(A \cap B)$ as $$P(A \cap B) = 0.6 + 0.7 - 1 = 0.3$$

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