# Probability of either of independent events

There are two independent events, $G$ and $H$. The probability that $G$ will occur is $r$ and the probability that $H$ will occur is $s$. I want to find the probability that either $G$ will occur or $H$ will occur, but not both!

I thought the answer would be r+s but that's not it. Can someone help me figuring where I might be wrong?

First, the definition of independence will allow you to compute $P(G \text{ and } H)$; what is it?

$P(G \text{ and } H) = P(G) \cdot P(H)$.

Next, check that the following equations are true.

$$P(\text{G occurs but not H}) = P(G) - P(G \text{ and } H).$$ $$P(\text{H occurs but not G}) = P(H) - P(G \text{ and } H).$$ $$P(\text{either G or H occurs but not both})=P(\text{G occurs but not H}) + P(\text{H occurs but not G}).$$

Combining the above facts yields the answer.

Since $G,H$ are independent, you know that $$\mathbb{P}[G \cap H] = \mathbb{P}[G] \cdot \mathbb{P}[H]$$ You have $$\mathbb{P}[G \cup H] = \mathbb{P}[G] + \mathbb{P}[H] - \mathbb{P}[G \cap H]$$ and you need $$\mathbb{P}[G \cup H] - \mathbb{P}[G \cap H].$$ Can you take it from here?

• So here P[G] is r(1-s)? – Sarvagya Gupta Nov 9 '17 at 8:10
• @SarvagyaGupta No, but $\mathbb{P}[G \cap H] = r \cdot s$. – gt6989b Nov 10 '17 at 0:26

Always consider the sample space... That is what are all the possibilities.

G and H both occurring has probability $rs$. G occurring and H not occurring has probability $r(1-s)$. G not occurring and H occurring has probability $(1-r)s$. Both G and H not occurring has probability $(1-r)(1-s)$.

You can check that these probabilities sum to 1.

The second and third cases are the ones we're interested in. Since they're mutually exclusive (they cannot happen at the same time) we can sum them.

The answer should be $r(1-s)+(1-r)s$.

The sum $r+s$ includes the case where both G and H occur (in fact that sum double counts that case; the case is included in both $r$ and $s$). You need to know the probability for both to occur - call that $t$. The probability for G or H but not both is then: $r+s-2t$.