# Suppose X has a binomial distribution

Question:

Suppose $X$ has a binomial distribution $B\left(6, \frac{1}{2}\right)$. Show that $X=3$ is the most likely outcome.

(Hint: P(X = 3) is the maximum among all P (xi), xi = 0, 1, 2, 3, 4, 5, 6)

Solution:

$X$ is the random variable whose binomial distribution is $B\left(6, \frac{1}{2}\right)$.

Therefore, $n=6$ and $p=\frac{1}{2}$

$\therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}$

Then, $\mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} q^{n-x} p^{x}$

$={ }^{6} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{6-x} \cdot\left(\frac{1}{2}\right)^{x}$

$={ }^{6} \mathrm{C}_{x}\left(\frac{1}{2}\right)^{6}$

It can be seen that $\mathrm{P}(\mathrm{X}=\mathrm{x})$ will be maximum, if ${ }^{6} \mathrm{C}_{x}$ will be maximum.

Then, ${ }^{6} \mathrm{C}_{0}={ }^{6} \mathrm{C}_{6}=\frac{6 !}{0 ! 6 !}=1$

${ }^{6} \mathrm{C}_{1}={ }^{6} \mathrm{C}_{5}=\frac{6 !}{1 ! 5 !}=$

${ }^{6} \mathrm{C}_{2}={ }^{6} \mathrm{C}_{4}=\frac{6 !}{2 ! 4 !}=15$

${ }^{6} \mathrm{C}_{5}=\frac{6 !}{3 ! \cdot 3 !}=20$

The value of ${ }^{6} \mathrm{C}_{3}$ is maximum. Therefore, for $x=3, \mathrm{P}(\mathrm{X}=\mathrm{x})$ is maximum.

Thus, X = 3 is the most likely outcome.