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As title states, I have 3 events X, Y, Z in a sample space each with a different probability of occurring, and I wanted to find the probability of exactly one of them happening. For reference, Z is independent of X and Y, while X and Y are dependent.

I know for just two events (say X and Y), I would use the formula: $P(X) + P(Y) - 2P(XY)$ Is there a similar way to solve for finding one of three events? Thanks.

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2 Answers

Hint: Add up the area you want to calculate! Venn

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So would this give me the correct answer: $P(A) + P(B) + P(C) - 2P(AB) - 2P(BC) - 2P(AC) - P(ABC)$ ? Or do I not need to subtract $P(ABC)$, since it's already covered in $2P(BC), 2P(AC), 2(AB)$? –  Smitty Shultz Apr 28 at 15:37
Well, how many times have you counted $ABC$? It's counted with a + in A, + in B, + in C, -2 in AB, -2 in BC, -2 in AC, and you want it counted 0 times overall. –  Sharkos Apr 28 at 21:31
So I should add 3(ABC) to the above equation and that'll give me 0 times overall, it seems, right? –  Smitty Shultz Apr 28 at 23:38
Sounds good to me! –  Sharkos Apr 29 at 8:05
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What you want should be:

$$ P(\text{Exactly one event})=P(X)+P(Y)+P(Z)-2P(X\cap Y)-2P(X\cap Z)-2P(Y\cap Z)+3P(X \cap Y \cap Z)=P(X)+P(Y)+P(Z)-2P(X\cap Y) $$ since $X, Y$ are independent of $Z$. You need to add $3P(X\cap Y\cap Z)$ since you subtract it six times ($2P(X\cap Y)$, $2P(X\cap Z)$ and $2P(Y\cap Z)$).

Edit: Corrected my answer. This time it should be correct...

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Even though Z is independent of Y and X, doesn't $P(XYZ)$ exist, and needs to be subtracted, along with $P(XZ)$ and $P(YZ)$? –  Smitty Shultz Apr 28 at 15:35
You are absolutely right. Removing my answer for now. Thanks! –  hejseb Apr 28 at 16:56
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