Question

The mean and variance of a binomial distribution are $4$ and $2$ respectively. Then the probability of $2$ successes is

• A. $\frac{28}{256}$
• B. $\frac{219}{256}$
• C. $\frac{128}{256}$
• D. $\frac{37}{256}$

Answer 1

The correct option is A

$\frac{28}{256}$

Explanation for the correct option.

Step 1: Find the probability of success and probability of failure.

Assume that, $p$ be the probability of success and $q$ be the probability of failure.

Therefore, $p+q=1...\left(1\right)$.

Also, the mean of a binomial distribution is given by $np$ and variance of a binomial distribution is given by $npq$.

Since, the mean of the given binomial is $4$.

Therefore, $np=4...\left(2\right)$.

Since, the variance of the given binomial is $2$.

$npq=2...\left(3\right)$

From equation $2$ and $3$.

$$\begin{gathered} 4q=2 \\ \Rightarrow q=\frac{1}{2} \end{gathered}$$

Therefore, the probability of failure is $\frac{1}{2}$.

From equation $1$.

$$\begin{gathered} p=1-\frac{1}{2} \\ \Rightarrow p=\frac{1}{2} \end{gathered}$$

Therefore, the probability of success is $\frac{1}{2}$.

Step 2: Find the probability of two successes.

Find the number of trials $n$.

Since, the probability of failure is $\frac{1}{2}$.

From equation $2$, we get

$$\begin{gathered} \frac{n}{2}=4 \\ \Rightarrow n=8 \end{gathered}$$

Therefore, the number of trials is $8$.

Find the probability $P$ of two successes as follows:

$$\begin{gathered} P={}^{8}C_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{8-2} \\ \Rightarrow P=28{\left(\frac{1}{2}\right)}^8 \\ \Rightarrow P=\frac{28}{256} \end{gathered}$$

Therefore, the probability $P$ of two successes is $\frac{28}{256}$.

Hence, option $A$ is the correct option.