Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# 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 - Mathematics

#### Question

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

• 32

• 34

• 33

• 31

#### Solution

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 (rnp) 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.
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#### APPEARS IN

RD Sharma Solution for Mathematics for Class 12 (Set of 2 Volume) (2018 (Latest))
Chapter 33: Binomial Distribution
MCQ | Q: 15 | Page no. 28

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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 Concept: Bernoulli Trials and Binomial Distribution.
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