Question

If $X$ follows a binomial distribution with parameter $n=101$ and $p=1/3$, then $P(X=r)$ is maximum if $r$ equals

• A. $34$
• B. $33$
• C. $32$
• D. $35$

Answer 1

Answer: (A), (B)

The correct options are
A $34$
B $33$

Given $n=101,p=\frac{1}{3}$
$\Rightarrow q=1-\frac{1}{3}=\frac{2}{3}$
We will check using options
$P(X=34){=}^{101}{C}_{34}\left(\frac{1}{3}{)}^{34}\right(\frac{2}{3}{)}^{67}$
$\Rightarrow P(X=34)=\frac{101}{\left(67\right)!\left(34\right)!}\times (\frac{1}{3}{)}^{101}{2}^{67}$ .....(1)
$P(X=33){=}^{101}{C}_{33}\left(\frac{1}{3}{)}^{33}\right(\frac{2}{3}{)}^{68}$
$\Rightarrow P(X=33)=\frac{101}{\left(68\right)!\left(33\right)!}\times (\frac{1}{3}{)}^{101}{2}^{68}$ .....(2)
$P(X=32){=}^{101}{C}_{32}\left(\frac{1}{3}{)}^{32}\right(\frac{2}{3}{)}^{69}$
$\Rightarrow P(X=32)=\frac{101}{\left(69\right)!\left(32\right)!}\times (\frac{1}{3}{)}^{101}{2}^{69}$ .....(3)
$P(X=35){=}^{101}{C}_{35}\left(\frac{1}{3}{)}^{35}\right(\frac{2}{3}{)}^{66}$
$\Rightarrow P(X=35)=\frac{101}{\left(66\right)!\left(35\right)!}\times (\frac{1}{3}{)}^{101}{2}^{66}$ .....(4)
Now, dividing (1) by (2), we get
$\frac{P(X=34)}{P(X=33)}=\frac{34}{68}\times 2=1$
$\Rightarrow P(X=34)=P(X=33)$
Now, we will divide (1) by (3)
$\frac{P(X=34)}{P(X=32)}=\frac{69}{64}>1$
$\Rightarrow P(X=34)>P(X=32)$
Now, we will divide (1) by (4), we get
$\frac{P(X=34)}{P(X=35)}=\frac{70}{67}>1$
$\Rightarrow P(X=34)>P(X=35)$
Hence, $P(X=34),P(X=33)$ is maximum.