Question

If the difference between the mean and variance of a binomial variate is \( \dfrac{5}{9} \), then the probability of getting 2 successes for an event when the experiment is conducted 5 times is:

• A. \(\dfrac{80}{243}\)
• B. \(\dfrac{18}{234}\)
• C. \(\dfrac{12}{241}\)
• D. \(\dfrac{80}{432}\)

Hint:

Use binomial mean and variance formulas: \( \mu = np \), \( \sigma^2 = np(1-p) \), and solve for \( p \) using given difference.

Answer 1

The Correct Option is A

Let \( X \sim \text{Bin}(n=5, p) \) Mean = \( np \), Variance = \( np(1 - p) \) Difference: \[ np - np(1 - p) = \dfrac{5}{9} \Rightarrow np[p] = \dfrac{5}{9} \] Since \( n = 5 \), \[ 5p^2 = \dfrac{5}{9} \Rightarrow p^2 = \dfrac{1}{9} \Rightarrow p = \dfrac{1}{3} \] Now, find \( P(X = 2) \) using binomial formula: \[ P(X = 2) = \binom{5}{2} \left(\dfrac{1}{3}\right)^2 \left(\dfrac{2}{3}\right)^3 = 10 \cdot \dfrac{1}{9} \cdot \dfrac{8}{27} = \dfrac{80}{243} \] % Final Answer \[ \boxed{P(X = 2) = \dfrac{80}{243}} \]