Toss a coin 3x, let $B = \text{Second flip is H}$ and $D =\space \ge \text{2 flips H}$, find $P(B | D)$
$P(B | D) = \frac{P(BD)}{P(D)}$, we know $P(D) = 1/2$, $P(B) = 1/2$.,
How do I get $P(BD)$? Applying complement, $P(BD) = 1 - P(\overline{B} \cup \overline{D})$
Fix the second throw, which must be heads. Now we can temporarily ignore that throw and consider the rest of the throws. We need to find the possible outcomes of $2$ throws s.t. at least one of them is head. Those are $HHT, THH, HHH$. In fact the number of outcomes from $n$ throws s.t. $k$ of them are heads is given by $\sum_{i=k}^{n} \binom{n}{i}$.
Therefore we can conclude that $P(B \cap D) = \frac{3}{8}$
$B\cap D$ is "second flip is heads and at least two flips are heads."
Which is "second flip is heads and not both of the other two are tails."
$$\tfrac 12\cdot(1-\tfrac 12\cdot\tfrac 12) = \tfrac 38$$