Editing Binomial distribution (section)

=== Conditional binomials ===
If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').

For example, imagine throwing ''n'' balls to a basket ''U<sub>X</sub>'' and taking the balls that hit and throwing them to another basket ''U<sub>Y</sub>''. If ''p'' is the probability to hit ''U<sub>X</sub>'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U<sub>X</sub>''. If ''q'' is the probability to hit ''U<sub>Y</sub>'' then the number of balls that hit ''U<sub>Y</sub>'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').

{{hidden begin|style=width:60%|ta1=center|border=1px #aaa solid|title=[Proof]}}
Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],
: <math>\begin{align}
   \Pr[Y = m] &= \sum_{k = m}^{n} \Pr[Y = m \mid X = k] \Pr[X = k] \\[2pt]
   &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}
 \end{align}</math>
Since <math>\tbinom{n}{k} \tbinom{k}{m} = \tbinom{n}{m} \tbinom{n-m}{k-m},</math> the equation above can be expressed as
: <math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math>
Factoring <math> p^k = p^m p^{k-m} </math> and pulling all the terms that don't depend on <math> k </math> out of the sum now yields
: <math>\begin{align}
   \Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]
   &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k}  \right)
 \end{align}</math>
After substituting <math> i = k - m </math> in the expression above, we get
: <math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math>
Notice that the sum (in the parentheses) above equals <math> (p - pq + 1 - p)^{n-m} </math> by the [[binomial theorem]]. Substituting this in finally yields
: <math>\begin{align}
   \Pr[Y=m] &=  \binom{n}{m} (pq)^m (p - pq + 1 - p)^{n-m}\\[4pt]
   &= \binom{n}{m} (pq)^m (1-pq)^{n-m}
 \end{align}</math>
and thus <math> Y \sim B(n, pq) </math> as desired.
{{hidden end}}