Suppose $X$ has a binomial distribution
$B\left( {6,\cfrac{1}{2}} \right)$ . Show that $X = 3$ is the most likely outcome. Hint: $P(X = 3)$ is the maximum among all $\left. {P\left( {{x_i}} \right),{x_i} = 0,1,2,3,4,5,6} \right)$

.: Given that $X$ has a binomial distribution $B\left( {6,\cfrac{1}{2}} \right)$

here $n = 6,p = \cfrac{1}{2}$

$\Rightarrow q = 1 - p = \cfrac{1}{2}$

Here ${(p + q)^n} = 3$

${\left( {\cfrac{1}{2} + \cfrac{1}{2}} \right)^6}{ = ^6}{C_0}{\left( {\cfrac{1}{2}} \right)^6}{ + ^6}{C_1}{\left( {\cfrac{1}{2}} \right)^6} + \ldots \ldots \ldots .{ + ^6}{C_6}{\left( {\cfrac{1}{2}} \right)^6}$

$= {\left( {\cfrac{1}{2}} \right)^6}{[^6}{C_0}{ + ^6}{C_1} + \ldots \ldots \ldots { + ^6}{C_6}]$

$= {\left( {\cfrac{1}{2}} \right)^6}{[^6}{C_0}{ + ^6}{C_1}{ + ^6}{C_2}{ + ^6}{C_3}{ + ^6}{C_2}{ + ^6}{C_1}{ + ^6}{C_0}]$

$6{C_3}$ has the maximum value in $^6{C_0}{ + ^6}{C_1}$ and $^6{C_2}$ .

Hence, $^6{C_3}{\left( {\cfrac{1}{2}} \right)^6}$

is maximum $\Rightarrow P(X = 3)$ is most likely outcome.