Which way to calculate this probability is correct? There is an urn that contains $N$ balls. Each ball might be either white or blue. I dont know how many white balls are in the urn, but my prior is that a ball is blue with probability $b$. Someone iterates through the blue balls and, for each ball independently, they either show it to me, with probability $p$, or they don't. What is the probability that I am shown $v$ balls?
Solution 1:
$Pr=\sum\limits_{i=0}^{N}{N\choose i}b^{i}(1-b)^{N-i}{i\choose v}p^{v}(1-p)^{i-v}.$
Solution 2:
$Pr={N \choose v}(bp)^v(1-bp)^{N-v}.$
I wish the second one is correct, but it must be wrong. Why?
 A: The formulas are equivalent:
$$\sum\limits _{i=v}^{N}{N \choose i}b^{i}(1-b)^{N-i}{i \choose v}p^{v}(1-p)^{i-v}=\binom{N}{v}\left(bp\right)^{v}\sum\limits _{i=0}^{N-v}\binom{N-v}{i}(b-bp)^{i}(1-b)^{N-v-i}=\binom{N}{v}\left(bp\right)^{v}\left(1-bp\right)^{N-v}$$
A: I understand why solution (1) is correct. Maybe you could give your argument for formula (2) so we could see where the reasoning is wrong (another possibility is that (1) and (2) are always equal by some combinatorial argument -- I didn't give that any thought yet EDIT: they're equivalent, see above answer).
At any rate, for now let's prove (1). For each $i$ with $0 \leq i \leq n$, suppose there are actually $i$ blue balls in the bin. We will compute the probabilities in each of these scenarios (call it "scenario $i$") and add them up. 
First of all, what is the probability that we are in scenario $i$, given our prior that the probability of a ball being blue is $b$? It is ${N \choose i}b^i(1-b)^{N-i}$.
Now, given that we are in scenario $i$, what is the probability that we see $v$ blue balls? It is ${i \choose v}p^v(1-p)^{i-v}$.
Since the scenarios $i$ are disjoint from one another as $i$ varies, we recover formula (1).
