Suppose X follows a binomial distribution with parameters n = 100 and $p=\frac{1}{3}.$Then P(X = r) is maximum when r = 

P(X=r) will be maximum when r is the mode. 

 If (n+1)p is not an integer there exists unique modal value and it's the integral part of (n+1)p. 

The given Binomial distribution has parameters, n=100 and p=1/3.  

We have (n+1)p=$(100+1)\frac{1}{3}=\frac{101}{3}$​=33.67, which is not an integer. 

Hence, the unique mode is 33, which is the integral part of (n+1)p.

Therefore, P(X=r) will be maximum when r=33
 

Given $\mathrm{n}=100, \mathrm{P}=\frac{1}{2} \Rightarrow \mathrm{q}=1-=\frac{1}{2}$

In Binomial distribution

$$\begin{gathered} \mathrm{P}(\mathrm{X}=\mathrm{r})={}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}} {\mathrm{p}}^{\mathrm{r}} {\mathrm{q}}^{\mathrm{n}-\mathrm{r}} \\ ={}^{100}{\mathrm{c}}_{\mathrm{r}} {\left(\frac{1}{2}\right)}^{\mathrm{r}} {\left(\frac{1}{2}\right)}^{100-\mathrm{r}} \end{gathered}$$

We know that ${}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}}$ is maximum at $\mathrm{r}=\frac{\mathrm{n}}{2}$

$\therefore {}^{100}{\mathrm{c}}_{\mathrm{r}}$ is maximum at $\mathrm{r}=\frac{100}{2}=50$

$\therefore \mathrm{P}(\mathrm{X}=\mathrm{r})$ is maximum when r=50