Q: Suppose 3 fair coins are flipped. What is the probability that all 3 are heads conditional on knowing the number of heads is odd? 
Suppose 3 fair coins are flipped. What is the probability that all 3 are heads conditional on knowing the number of heads is odd?

The answer is $1/4$ and I don't understand how.
Using $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$
The probability of having 3 heads is $\frac{1}{8}$, the probability of having an odd number of heads is $\frac{1}{2}$
So $P(A\mid B)=\frac{(1/8)(1/2)}{(1/2)}=1/8$ What am I not understanding?

A: Hint:
$Pr(A\cap B) = Pr(A)\times Pr(B)$ is true if and only if $A$ and $B$ are independent events.  These events certainly are not independent and so the above would be an inequality, not an equality.
Instead, recognize that "All three are heads" directly implies that "The number of heads is odd"

 And so $A\subseteq B$ in this specific situation.

$~$

 And so $Pr(A\cap B)=Pr(A)$ in this specific situation.

