Why $P((A \cup B) \cap C) \neq P((A \cap C) \cup (B \cap C))$?

Question

Let's consider random independent events A, B, C with $P(A)=0.2$, $P(B)=0.3$ and $P(C)=0.4$. Then $P((A \cup B) \cap C)=P(A \cup B)P(C)=0,176$. I would also expect that $P((A \cup B) \cap C)=P((A \cap C) \cup (B \cap C))$. However, $P((A \cap C) \cup (B \cap C))=0,1904$. Why doesn't this approach work?

Answer 1

Given that $A,B,C$ are mutually independent:

$P((A\cup B)\cap C)=P(A\cup B)P(C)=(P(A)+P(B)-P(A\cap B))P(C)=(P(A)+P(B)-P(A)P(B))P(C)=$

$=(0.2+0.3-0.2\cdot 0.3)\cdot 0.4=0.176$

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$P((A\cap C)\cup (B\cap C))=P(A\cap C)+P(B\cap C)-P((A\cap C)\cap (B\cap C))$

$=P(A)P(C)+P(B)P(C)-\underbrace{P(A\cap B\cap C)}$

$0.2\cdot 0.4+0.3\cdot0.4-0.2\cdot0.3\cdot0.4=0.176$

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You appear to have calculated $P((A\cap C)\cap (B\cap C))$ as $P(A)P(C)P(B)P(C)$ instead. $A\cap C$ is not independent from $B\cap C$. (alternatively phrased, $C$ is not independent from $C$)

Answer 2

Apparently you calculated the value of $P((A\cap C) \cup (B\cap C))$ using the formula that only applies for independent events. However, $A\cap C$ and $B\cap C$ are not independent.