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There are 3 events $A$, $B$, and $C$ that are mutually independent and $P(A) =$ the probability of event $A$ occurring $= P(B) = P(C) = 0.05$. Let's say that $\cap$ represents the intersection symbol. For example, $A\cap B = A$ intersection $B$. I need to find $P(A\cap B'\cap C')$.

I first thought about using the rules of mutually independent events. Since, A,B, and $C$ are mutually independent, $P(A\cap B\cap C) = P(A)P(B)P(C)$. Also, I know that $A$ and $(B\cap C)$ are mutually independent, $A$ and $(B \cup C)$ are mutually independent, $A'$ and $(B \cap C')$ are mutually independent, and $A'$, $B'$, and $C'$ are mutually independent. However, I do not know if $A$, $B'$, and $C'$ are mutually independent which is what appears in the probability I am looking for.

I then tried to rewrite the probability I am looking for in a different way by saying that $$P(A\cap B'\cap C') = 1 - (P(A\cap B'\cap C')') = 1 - P(A' \cup B \cup C).$$ I then tried to rewrite the above probability by saying $$1 - P(A' \cup B \cup C) = 1 - ((P(A) + P(B) + P(C) - P(A'\cap B) - P(A'\cap C) - P(B\cap C) + P(A'\cap B\cap C)).$$ All but the last term could be solved for by using the given probability and the rules of mutually independent events. However, the last term is again in a state where I do not know if the terms are mutually independent, so I cannot think of a way to solve for it. I am a bit stumped at this point because I'm not able to think of another way to rewrite the probability I need to find in a useful way.

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  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Sep 8 '18 at 20:08
  • $\begingroup$ Given that $A_1,A_2,A_3,\dots,A_n$ are mutually independent so too are then any combination of these where any number of them are replaced by their complements instead. Yes $P(A\cap B'\cap C')=P(A)(1-P(B))(1-P(C))$ for mutually independent events $A,B,C$. $\endgroup$ – JMoravitz Sep 8 '18 at 20:21
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Using $P(A\cap B')=P(A)-P(A\cap B)$: $$\begin{align} P(A \cap B' \cap C')&=P(A \cap (B \cup C)')\\ &=P(A)-P(A\cap (B\cup C))\\ &=P(A)-P((A\cap B)\cup (A\cap C))\\ &=P(A)-(P(A\cap B)+P(A\cap C)-P(A\cap B\cap C))\\ &=P(A)(1-P(B)-P(C)-P(B)P(C))\\ &=P(A)(1-P(B))(1-P(C)) \end{align}$$

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