# If the Mean and Variance of a Binomial Distribution Are 4 and 3, Respectively, the Probability of Getting Exactly Six Successes in this Distribution is - Mathematics

MCQ

If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is

#### Options

• $^{16}{}{C}_6 \left( \frac{1}{4} \right)^{10} \left( \frac{3}{4} \right)^6$

• $^{16}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}$

• $^{12}{}{C}_6 \left( \frac{1}{20} \right) \left( \frac{3}{4} \right)^6$

• $^{12}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^6$

#### Solution

$^{16}{}{C}_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}$

Mean (np) = 4 and Variance (npq) = 3

$\therefore q = \frac{3}{4}$
$\Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}\text{ and } n = 16$
$\text{ Let X denotes the number of successes in 16 trials . Then, }$
$P(X = r) = ^{16}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{16 - r}$
$\Rightarrow P(X = 6) = \text{ Probability (getting exactly 6 successes } )$
$= 16 C_6 \left( \frac{1}{4} \right)^6 \left( \frac{3}{4} \right)^{10}$

Concept: Bernoulli Trials and Binomial Distribution
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 33 Binomial Distribution
MCQ | Q 20 | Page 29