Question

If $x$ follows a binomial distribution with parameters $n=100$ and $p=\frac{1}{3}$, then $P(x-r)$ is maximum, when $r$ equals

• A. $16$
• B. $32$
• C. $33$
• D. None of these

Answer 1

Answer: (C)

$33$

The binomial distribution is given by,

$f(r,n,p)=\rho (X=r){=}^n{C}_r{p}^r{(1-p)}^{n-r}$

This value can be maximum at a particular r, which can be determined as follows,

$\frac{f(r+1,n,p)}{f(r,n,p)}=1$

$\Rightarrow \frac{{}^n{C}_{r+1}{\left(p\right)}^{r+1}\times {(1-p)}^{n-r-1}}{{}^n{C}_r{\left(p\right)}^r\times {(1-p)}^{n-r}}=\frac{(n-r)p}{(r+1)(1-p)}=1$

On Substituting the values of $n=100,p=\frac{1}{3}$ , we get

$\frac{(100-r)\times \frac{1}{3}}{(r+1)(1-\frac{1}{3})}=1$

$\Rightarrow (100-r)\frac{1}{3}=(r+1)\frac{2}{3}$

$\Rightarrow (100-r)=(r+1)2$

$\Rightarrow 100-r=2r+2$

$\Rightarrow 98=3r$

$\therefore r=\frac{98}{3}$

The integer value of r satisifes $(n+1)p-1\le m\le (n+1)p$

$f(r,n,p)$ is monotonically increasing for r<m and monotonically decreasing from $r>m$

as $\frac{98}{3}\le m<\frac{101}{3}$

$\therefore$ The integer value of r is $33.$