The question: Jay and John each flip fair coins until they obtain their first heads, respectively. Given that it takes strictly fewer flips for Jay to get his first heads than John, compute the expected number of flips Jay performed.
I set up the problem as follows: X = the number of flips it takes Jay to get heads, Y for John. Then the problem becomes $E[X | X < Y] = \sum_{k=1}^{\infty} kP(X = k | X<Y)$.
This is equal to $\sum_{k=1}^{\infty} k\dfrac{P(X = k, X<Y)}{P(X<Y)}$.
We know that X, Y are independent. Why is it invalid to then say that $\sum_{k=1}^{\infty} k\dfrac{P(X = k, X<Y)}{P(X<Y)} = \sum_{k=1}^{\infty} k\dfrac{P(X = k)P(X<Y)}{P(X<Y)}$?
The solution suggests writing the below instead:
