Question

How do I solve this problem?

Answer 1

The probability you're asked to find is

$\mathbb P(A\cup B)-\mathbb P(A\cap B)$

where $A\cup B$ is the event that either event occurs (A, B, or both), and $A\cap B$ is the event that both events occur.

Recall that

$\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)$

which means

$\mathbb P(A\cup B)-\mathbb P(A\cap B)=\mathbb P(A)+\mathbb P(B)-2\mathbb P(A\cap B)$

You're told that $\mathbb P(A)=0.5$, $\mathbb P(B)=0.7$, and (if I'm reading the diagram correctly) $\mathbb P(A\cap B)=0.3$.

So,

$\mathbb P(A\cup B)-\mathbb P(A\cap B)=0.5+0.7-2\times0.3=0.6$

Another way of seeing this is that the event A consists of the regions "A not B" and "A and B". So the probability that "A not B" occurs is

$\mathbb P(A\setminus B)=\mathbb P(A)-\mathbb P(A\cap B)=0.5-0.3=0.2$

Similarly, B consists of "B not A" and "A and B", so you have

$\mathbb P(B\setminus A)=\mathbb P(B)-\mathbb P(A\cap B)=0.7-0.3=0.4$

So the probability that A or B, but not both, occur is

$\mathbb P(A\setminus B)+\mathbb P(B\setminus A)=0.2+0.4=0.6$