Question

If X follows a binomial distribution with parameters n = 100 and p = 1/3, then P (X = r) is maximum when r =

Options

• 32

• 34

• 33

• 31

   

Answer 1

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,

\[\frac{f\left( r + 1, n, p \right)}{f\left( r, n, p \right)} = 1\]

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

On substituting the values of n = 100,

\[p = \frac{1}{3}\]  ,

we get , 

\[\frac{\left( 100 - r \right) \times \frac{1}{3}}{\left( r + 1 \right) \left( 1 - \frac{1}{3} \right)} = 1\]
\[ \Rightarrow \left( 100 - r \right)\frac{1}{3} = \left( r + 1 \right)\frac{2}{3}\]
\[ \Rightarrow 100 - r = \left( r + 1 \right)2\]
\[ \Rightarrow 100 - r = 2r + 2\]
\[ \Rightarrow 98 = 3r\]
\[ \Rightarrow 3r = 98\]
\[ \therefore r = \frac{98}{3}\]

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

\[\text{ as } \frac{98}{3} \leq m < \frac{101}{3}\]

∴ The integer value of r is 33.