# Calculating the probability of an event giving the union and complement

There are two independent events. The probability that both occurs at the same time is $\frac{1}{6}$ and the probability that none of them happens is $\frac{2}{3}$. What is the probability that only one of them occurs?

I'm trying to solve it but I cannot find a way to solve it to just one event. Here's what I did:

We know that $P(A\cap B) = \frac{1}{6}$ and $[1 - P(A)] \cdot [1 - P(B)] = \frac{2}{3}$

So,

$$[1 - P(A)] \cdot [1 - P(B)] = \frac{2}{3}$$ $$1 - P(A) - P(B) + P(A) \cdot P(B) = \frac{2}{3}$$ $$1 - P(A) - P(B) + \frac{1}{6}= \frac{2}{3}$$

As can be seen, I cannot can solve for just $P(A)$ or just $P(B)$. Can someone give me a hint what a I'm doing wrong? I already know that the answer for this question is $\frac{1}{6}$

• both occurs at same time should be "intersection" not union. For independent event that will be equal to $P(A)* P(B)$ May 22, 2016 at 3:30
• That's right! Good looking out May 22, 2016 at 3:33
• Second part is also wrong.. that will be complement of $P(A U B) =2/3$ May 22, 2016 at 3:36
• @ViX28 The second part is ok. You can double check with the identity, valid for all events $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$
– user228113
May 22, 2016 at 3:41
• exactly..my mistake. Its same. May 22, 2016 at 3:44

$$\mathsf P(A\oplus B) = \mathsf P(A\cup B)-\mathsf P(A\cap B)$$
When you know $$\mathsf P(A\cap B)=1/6 \\ 1-\mathsf P(A\cup B) = 2/3$$
• So we have a negative probability? It's impossible. P(A$\cup$B) = $\frac{2}{3}$ - 1 = -$\frac{1}{3}$ May 26, 2016 at 23:06
• @Rods2292 $1-p=x$ means $p=1-x$ May 26, 2016 at 23:37
Hint 2: It's actually impossible in this question to find either probability of $A$ or of $B$ individually, but fortunately that's not what the question is asking. It's asking for the probability that only one event occurs, not both simultaneously, but it doesn't specify which one. Using notation, it's asking for $P(A\cap\neg B)+P(B\cap\neg A)$. And you can easily see that from the diagram.