Suppose we have $A \sim Poisson (\lambda)$ and $B \sim Geometric(p)$, where $A$ and $B$ are independent. Calculate $P(A=a \mid A <B)$
My attempt: \begin{align*} P(A=a \mid A<B)&=\frac{P((A=a) \cap (A<B))}{P(A<B)}\\ &=\frac{P(A=a) \cdot P(a<B)}{P(A<B)} \mbox{ (by independence})\\ &=\frac{\frac{e^{-\lambda} \lambda^{a}}{a!} \cdot (1-p)^{a}}{P(A<B)} \end{align*}
So how I know how to calculate the denominator $P(A<B)$. However, I'm unsure if my numerator is correct, it just a somewhat lengthy calculation I do not show here. $\textbf{Should the numerator be just $P(a<B)$ or is mine correct with $P(A=a) \cdot P(a<B)$?}$
For example, if I was asked to calculate $P(A<B \mid B=b)$, $\textbf{would the numerator of this probability be $P(A<b)$ or would it be $P(A<b) \cdot P(B=b)$?}$ Thank you!
The point is that in the first one the event $A=a \cap A<B$ is not just $a<B$. It is however $A=a \cap a<B$, which doesn't result in any nice cancellation with the denominator in the conditional probability.
In the second one you again have $A<B \cap B=b=A<b \cap B=b$, but now you do get some cancellation because $P(B=b)$ in the numerator will cancel with $P(B=b)$ in the denominator. The difference was that the condition only involved one of the variables in the second case.