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Given $\Pr(A) = 0.7$ and $\Pr(B) = 0.9$, what is the highest possible value of $\Pr(A \cup B) - \Pr(A \cap B)$ ?

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  • Draw the Venn diagram. Done. You can find this approach explained and illustrated at https://stats.stackexchange.com/questions/466434. – whuber Aug 25 '22 at 13:50

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Note that $\Pr(A\cup B) \in [0,1] $. So if we assume that $\Pr(A\cup B) =1$, then we would have $\Pr (A\cap B) = \Pr(A) + \Pr(B) - \Pr(A\cup B) = 0.7 + 0.9 - 1 = 0.6$. Hence the biggest value $\Pr(A \cup B) - \Pr(A \cap B)$ will be $\Pr(A \cup B) - \Pr(A \cap B) = 1 - 0.6 = 0.4$

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    That is the highest possible value but not a proof. Perhaps say $\Pr(A \cup B)=\Pr(A ) +\Pr(B ) - \Pr(A \cap B)$ so $\Pr(A \cup B) - \Pr(A \cap B)=2\Pr(A \cup B) -\Pr(A ) -\Pr(B )$ which is maximised here when $\Pr(A \cup B)$ is maximised. And $\Pr(A \cup B)=1$ is the highest possible value here – Henry Aug 25 '22 at 08:02
  • @Henry +1 but note that the question does not ask for a proof, only for the highest possible value. – Dilip Sarwate Aug 25 '22 at 13:31