Question

If X follows a binomial distribution with parameters n = 100 and p = 1/3, then P (X = r) is maximum when r =
(a) 32
(b) 34
(c) 33
(d) 31

Answer 1

(c) 33

The binomial distribution is given by,
$f\left(r,n,p\right)=P\left(X=r\right)={}^{n}c_r{p}^r{\left(1-p\right)}^{n-r}$
This value can be maximum at a particular r, which can be determined as follows,
$$\begin{gathered} \frac{f\left(r+1,n,p\right)}{f\left(r,n,p\right)}=1 \\ \Rightarrow \frac{{}^{n}c_{r+1}{\left(p\right)}^{r+1}\times {\left(1-p\right)}^{n-r-1}}{{}^{n}c_r{\left(p\right)}^r\times {\left(1-p\right)}^{n-r}}=\frac{\left(n-r\right)p}{\left(r+1\right)\left(1-p\right)}=1 \end{gathered}$$
On substituting the values of n = 100, $p=\frac{1}{3}$, we get
$$\begin{gathered} \frac{\left(100-r\right)\times {\displaystyle \frac{1}{3}}}{\left(r+1\right)\left(1-{\displaystyle \frac{1}{3}}\right)}=1 \\ \Rightarrow \left(100-r\right)\frac{1}{3}=\left(r+1\right)\frac{2}{3} \\ \Rightarrow 100-r=\left(r+1\right)2 \\ \Rightarrow 100-r=2r+2 \\ \Rightarrow 98=3r \\ \Rightarrow 3r=98 \\ \therefore r=\frac{98}{3} \end{gathered}$$
The integer value of r satisfies (n + 1)p − 1 ≤ m < (n + 1)p

f (r, n, p) is montonically increasing for r < m and montonically decreasing for r > m

$\mathrm{as}\hspace{0.17em}\hspace{0.17em}\frac{98}{3}\le m<\frac{101}{3}$

∴ The integer value of r is 33.