Question

If the mean and variance of a binomial distribution are 4 and 2 respectively, then probability of getting 2 successes is

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

Hint:

Whenever the variance is exactly half of the mean ($npq = \frac{1}{2}np$), the distribution is perfectly symmetric, meaning $p = q = \frac{1}{2}$. This simplifies the calculation because the exponential term is always constant: $p^k q^{n-k} = \left(\frac{1}{2}\right)^n$. The final numerator is simply the combination value $\binom{n}{k}$!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
For a binomial distribution with mean 4 and variance 2, find P(X=2).

Step 2: Key Formula or Approach:
Mean=np, Variance=npq; solve for n and p, then use binomial probability formula.

Step 3: Detailed Explanation:
np=4, npq=2 → q=2/4=1/2 → p=1/2 → n=8. P(X=2) = C(8,2)(1/2)²(1/2)⁶ = 28 × (1/256) = 28/256.

Step 4: Final Answer:
P(X=2) = 28/256, matching option (A).