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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$\begingroup$Hint: Add up the area you want to calculate!
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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