Editing Binomial distribution (section)

=== Sums of binomials ===
If {{math|''X'' ~ B(''n'', ''p'')}} and {{math|''Y'' ~ B(''m'', ''p'')}} are independent binomial variables with the same probability {{math|''p''}}, then {{math|''X'' + ''Y''}} is again a binomial variable; its distribution is {{math|1=''Z'' = ''X'' + ''Y'' ~ B(''n'' + ''m'', ''p'')}}:<ref>{{cite book |last1=Dekking |first1=F.M. |last2=Kraaikamp |first2=C. |last3=Lopohaa |first3=H.P. |last4=Meester |first4=L.E. |title=A Modern Introduction of Probability and Statistics |date=2005 |publisher=Springer-Verlag London |isbn=978-1-84628-168-6 |edition=1 |url=https://www.springer.com/gp/book/9781852338961}}</ref>
: <math>\begin{align}
  \operatorname P(Z=k) &= \sum_{i=0}^k\left[\binom{n}i p^i (1-p)^{n-i}\right]\left[\binom{m}{k-i} p^{k-i} (1-p)^{m-k+i}\right]\\
                       &= \binom{n+m}k p^k (1-p)^{n+m-k}
\end{align}</math>

A Binomial distributed random variable {{math|''X'' ~ B(''n'', ''p'')}} can be considered as the sum of {{math|''n''}} Bernoulli distributed random variables. So the sum of two Binomial distributed random variables {{math|''X'' ~ B(''n'', ''p'')}} and {{math|''Y'' ~ B(''m'', ''p'')}} is equivalent to the sum of {{math|''n'' + ''m''}} Bernoulli distributed random variables, which means {{math|1=''Z'' = ''X'' + ''Y'' ~ B(''n'' + ''m'', ''p'')}}. This can also be proven directly using the addition rule.

However, if {{math|''X''}} and {{math|''Y''}} do not have the same probability {{math|''p''}}, then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as {{math|B(''n'' + ''m'', {{overline|''p''}})}}.