Computing probabilities of events

Question

Does anyone know how to solve this exercise?

Let A and B be two independent events.
If we additionally know that P (A|B) = 0.6 and P (B|A) = 0.3, compute the probabilities of the following events

• at most one of A or B
• either A or B but not both.

I know that $ P(A|B) = \frac{P(A \cap B)}{P(B)}$ and that independent means that $ P(A|B) = P(A) $, but I really have no idea how to get the solution using these formulas.

Answer 1

"At most one of $A$ or $B$" can be reworded as "Not the case it is simultaneously both $A$ and $B$"

$1 - \Pr(A\cap B)$

"Either $A$ or $B$ but not both"

$\Pr(A\cup B)-\Pr(A\cap B)$ or equivalently $\Pr(A\cap B^c) + \Pr(A^c\cap B)$

Now, the problem statement told you reworded that $\Pr(A)=0.6$, that $\Pr(B)=0.3$ and that $A$ and $B$ are independent. Recall that implies that $\Pr(A\cap B) = \Pr(A)\Pr(B)$, that $\Pr(A\cap B^c)=\Pr(A)(1-\Pr(B))$ and so on...

Recall also other common properties like $\Pr(A^c)=1-\Pr(A)$ and $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$