Find the probability that only one event happens

Question

$P(A) = 0.35$, $P(B) = 0.72$ Probability that at least one of these two events happen = $0.90$. What is the probability of only one of these events happen?

So far, I tried doing $P(A \cap B^c) + P(A^c \cap B)$ And also tried $1 - P(A \cap B) - P(A^c \cap B^c)$ but to no avail.

Answer 1

Hint: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ Then, since $(A\cap B)\subset (A\cup B)$, we have $$P((A\cup B)\text{ \ }(A\cap B))$$$$=P(A\cup B)-P(A\cap B)$$$$=2P(A\cup B)-P(A)-P(B)$$ $$=2\cdot0.90-0.35-0.72=0.73$$

Answer 2

You have $P(A), P(B), P(A\cup B)$ and should know that $\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)$ from the Principle of Inclusion and Exclusion.

Now find $\mathsf P((A\cup B)\smallsetminus(A\cap B))$, the probability for one or the other happening but not both.

Answer 3

hint...First use $p(A\cup B)=p(A)+p(B)-p(A\cap B)$ to get $p(A\cap B)$

Then you need $p(A\cup B)-p(A\cap B)$