Just as the title asks, I can't seem to wrap my head around a situation in which that is true. Of course I know that since A and B and A and C are independent, that P(A∩B)=P(A)P(B) and P(A∩C)=P(A)P(C). So what would be an example in which P(A∩(B∪C)) is not equal to P(A)P(B∪C)?
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See https://stats.stackexchange.com/a/180729/919. – whuber Feb 28 '21 at 21:18
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Let's say we have two fair coins, and the following events:
$A$: First coin is heads, $B$: second coin is heads and $C:$ only one of the coins is heads.
These event definitions satisfy the first two conditions. However,
$$P(A\cap (B\cup C))=1/2$$
while $P(A)P(B\cup C)=1/2 \times 3/4=3/8$.

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