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A possibility space is made up of four elementary events A, B, C, D where

$P(A) = p^2$

$P(B) = pq$

$P(C) = pq$

$P(D) = q^2$

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up vote 7 down vote accepted

We begin by noting that for a discrete probability space $\Omega$ we have:

$$\sum_{\omega \in \Omega} P(\omega) = 1$$

And $P(\omega) \in [0,1] \space\forall \omega \in \Omega$, therefore we have:

$$p^{2}+2pq+q^{2}=1 \implies (p+q)^{2}=1\implies p+q=1$$

We wish to find: $P(A \cup C)= P(A)+P(C)-P(A \cap C)$, but we note that $A, B, C$ and $D$ are elementary and thus $P(A \cap C) = 0$.

We therefore have that:

$$P(A \cup C) = p^{2}+pq=p(p+q)=p$$

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Thanks. This makes sense to me :) – StephanCasey Aug 10 '14 at 14:35
That's a great answer! – user159870 Aug 10 '14 at 14:37

Maybe this one would help: $$P(A \cup C)=P(A)+P(C)-P(A \cap C)=p^2+pq-P(A \cap C)$$

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