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How do I derive this expression for conditional probability?

Problem Statement
"Let ##Y_i## be independent Bernoulli r.v.'s. Let ##T=1## if ##Y_1=1## and ##Y_2=0## and ##T=0## otherwise. Let ##W=\sum Y_i##. Prove that ##P(T=1|W=w)=\frac{w(n-w)}{n(n-1)}##.
Relevant Equations
##P(A|B)=\frac{P(A\cap B)}{P(B)}##
An answer from another part in the problem: ##E(T)=p(1-p)##
##P(T=1|W=w)=\frac{P(\{T=1\}\cap\{W=w\})}{P(W=w)}=\frac{\binom {n-2} {w-1} p^{w-1}(1-p)^{(n-2)-(w-1)}}{\binom n w p^w (1-p)^{n-w}}=\frac{(n-2)!}{(w-1)!(n-w-1)!}\frac{w!(n-w)!}{n!}\frac{1}{p(1-p)}=\frac{w(n-w)}{n(n-1)}(p(1-p))^{-1}##.

I cannot seem to get the terms with ##p## out of my expression.
 

Math_QED

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Problem Statement
"Let ##Y_i## be independent Bernoulli r.v.'s. Let ##T=1## if ##Y_1=1## and ##Y_2=0## and ##T=0## otherwise. Let ##W=\sum Y_i##. Prove that ##P(T=1|W=w)=\frac{w(n-w)}{n(n-1)}##.
Relevant Equations
##P(A|B)=\frac{P(A\cap B)}{P(B)}##
An answer from another part in the problem: ##E(T)=p(1-p)##

##P(T=1|W=w)=\frac{P(\{T=1\}\cap\{W=w\})}{P(W=w)}=\frac{\binom {n-2} {w-1} p^{w-1}(1-p)^{(n-2)-(w-1)}}{\binom n w p^w (1-p)^{n-w}}=\frac{(n-2)!}{(w-1)!(n-w-1)!}\frac{w!(n-w)!}{n!}\frac{1}{p(1-p)}=\frac{w(n-w)}{n(n-1)}(p(1-p))^{-1}##.

I cannot seem to get the terms with ##p## out of my expression.
Your second equality is wrong.

For ##w \geq 1##, we have:

##P(T=1, W = w) = P(Y_1 = 1, Y_2 = 0, \sum_{i=3}^n Y_i = w-1) = P(Y_1 = 1)P(Y_2 = 0)P(\sum_{i=3}^n Y_i = w-1)## (here the independence of ##(Y_n)_n## was crucial) and the first two factors give the factor ##p(1-p)## that you are missing.

For ##w=0##, it is easily checked that the equality you want to prove holds as well.
 
Okay, thanks.
 

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