Question

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

Answer 1

$X$ is the random variable whose binomial distribution is $B(6,\frac{1}{2})$.
Therefore, $n=6$ and $p=\frac{1}{2}$
$\therefore q=1-p=1-\frac{1}{2}$
$=\frac{1}{2}$
Then, $P(X=x){=}^n{C}_x{q}^{n-x}{p}^x$
${=}^6{C}_x{\left(\frac{1}{2}\right)}^{6-x}\cdot {\left(\frac{1}{2}\right)}^x$
${=}^6{C}_x{\left(\frac{1}{2}\right)}^6$
It can be seen that $P(X=x)$ will be maximum, if ${}^6{C}_x$ will be maximum.
Then, ${}^6{C}_0{=}^6{C}_0=\frac{6!}{0!\cdot 6!}=1$
${}^6{C}_1{=}^6{C}_5=\frac{6!}{1!\cdot 5!}=6$
${}^6{c}_2{=}^6{C}_4=\frac{6!}{2!\cdot 4!}=15$
${}^6{C}_3=\frac{6!}{3!\cdot 3!}=20$
The value of ${}^6{C}_3$ is maximum. Therefore, for $x=3,P(X=3)$ is maximum.
Thus, $X=3$ is the most likely outcome.