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


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

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 $\frac{1}{6}$

  • $\begingroup$ both occurs at same time should be "intersection" not union. For independent event that will be equal to $P(A)* P(B)$ $\endgroup$
    – ViX28
    May 22, 2016 at 3:30
  • $\begingroup$ That's right! Good looking out $\endgroup$
    – Rods2292
    May 22, 2016 at 3:33
  • $\begingroup$ Second part is also wrong.. that will be complement of $ P(A U B) =2/3$ $\endgroup$
    – ViX28
    May 22, 2016 at 3:36
  • 1
    $\begingroup$ @ViX28 The second part is ok. You can double check with the identity, valid for all events $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ $\endgroup$
    – user228113
    May 22, 2016 at 3:41
  • $\begingroup$ exactly..my mistake. Its same. $\endgroup$
    – ViX28
    May 22, 2016 at 3:44

2 Answers 2


Hint: The probability that only one occurs is:

$$\mathsf P(A\oplus B) = \mathsf P(A\cup B)-\mathsf P(A\cap B)$$

When you know $$\mathsf P(A\cap B)=1/6 \\ 1-\mathsf P(A\cup B) = 2/3$$

  • $\begingroup$ So we have a negative probability? It's impossible. P(A$\cup$B) = $\frac{2}{3}$ - 1 = -$\frac{1}{3}$ $\endgroup$
    – Rods2292
    May 26, 2016 at 23:06
  • $\begingroup$ @Rods2292 $1-p=x$ means $p=1-x$ $\endgroup$ May 26, 2016 at 23:37

Hint 1: Draw a Venn diagram.

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.


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