Question

In a Binomial distribution, the mean is three times its variance. What is the probability of exactly $3$ successes out of $5$ trials?

• A. $\frac{80}{243}$
• B. $\frac{40}{243}$
• C. $\frac{20}{243}$
• D. $\frac{10}{243}$

Answer 1

Answer: (A)

$\frac{80}{243}$

Given that, in a binomial distribution the mean is three times its variance.

That is $np=3npq$

$\Rightarrow q=\frac{1}{3}$

Therefore $p=1-q=1-\frac{1}{3}=\frac{2}{3}$

$P(x=3)=5C_3{\left(\frac{2}{3}\right)}^3\times {\left(\frac{1}{3}\right)}^2$

$=10\times \frac{8}{27}\times \frac{1}{9}$

$\therefore P(x=3)=\frac{80}{243}$