Two events $A$ and $B$ are such that $P(A) = 9/16$, $P(B) = 3/8$, and $P(A|B) = 1/4$. For my IB Math class, I have a test on probability and statistics next week. I'm reviewing practice problems in the book and working on some to prepare myself for the test.    This is a problem I'm having a little trouble with: 
Two events $A$ and $B$ are such that $P(A) = 9/16$, $P(B) = 3/8$, and $P(A|B) = 1/4$. Find the probability that: 
a. both events happen 
b. only one of the events will happen
c. neither event will happen
for a, do I just multiply $P(A)$ and $P(B)$?
for b, is it $1/2$?
and I'm not entirely sure how c works. Thank you!
 A: "For (a), do I just multiply $P(A)$ and $P(B)$?"
No.
It is unfortunately common for people to learn and remember the rules for calculating the probability of intersection and unions incorrectly.
$P(A\cap B)=P(A)\times P(B)$ is true if and only if $A$ and $B$ are independent events.  In the case that we do not know that $A$ and $B$ are independent events, we may NOT just break it up like this.  In the case that $A$ and $B$ are not independent, in fact we have $P(A\cap B)\neq P(A)\times P(B)$.
Instead, something that is always true is the "multiplication principle of probability" that $P(A\cap B)=P(A)\times P(B\mid A)$ and similarly that $P(A\cap B)=P(B)\times P(A\mid B)$.  Your problem gives you exactly the information you need then to calculate $P(A\cap B)$.
Now, you can go further and calculate $P(A\cup B)$ using inclusion-exclusion which implies $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and note that the probability that exactly one of them occurring is going to be $P(A\cup B)-P(A\cap B)$.  You again have enough information to calculate this.
Finally, the probability that neither event occurs will be $1-P(A\cup B)$, which again you have enough information to calculate.
A: *

*$A$ and $B$ are not given as independent events, so you can't just multiply $P(A)$ and $P(B)$: $P(A|B)=\frac{P(A\cap B)}{P(B)}$ by definition, so $P(A\cap B)=P(A|B)P(B)=\frac{3}{32}$.

*Use the fact that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ (maybe it's a good exercise for you to prove this).

*Neither event happens means that both $A^c$ and $B^c$ happen, and we know that $A^c\cap B^c=(A\cup B)^c$ by De Morgan's law, so use $P((A\cup B)^c)=1-P(A\cup B)$.
