Solution:

Given $\mathrm{X}$ is any random variable whose binomial distribution is $\mathrm{B}(6,1 / 2)$

Thus, $\mathrm{n}=6$ and $\mathrm{p}=1 / 2$

$q=1-p=1-1 / 2=1 / 2$

Thus, $P(X=x)={ }^{n} C_{x} q^{n-x} p^{x}$, where $x=0,1,2 \ldots n$

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

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

It can be clearly observed that $\mathrm{P}(\mathrm{X}=\mathrm{x})$ will be maximum if ${ }^{6} \mathrm{c}_{\mathrm{x}}$ will be maximum.

$$\therefore{ }^{6} \mathrm{c}{\mathrm{x}}={ }^{6} \mathrm{C}{6}=1$$

$${ }^{6} \mathrm{c}{1}={ }^{6} \mathrm{c}{5}=6$$

$${ }^{6} \mathrm{c}{2}={ }^{6} \mathrm{C}{4}=15$$

$${ }^{6} \mathrm{C}_{3}=20$$

Hence we can clearly see that ${ }^{6} \mathrm{c}_{3}$ is maximum.

$\therefore$ for $x=3, P(X=x)$ is maximum.

Hence, proved that the most likely outcome is $x=3$.