Question

If the mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then \( P(X=2)= \)

• A. \( \frac{7}{64} \)
• B. \( \frac{15}{64} \)
• C. \( \frac{21}{64} \)
• D. \( \frac{39}{64} \)

Hint:

Whenever the success and failure probabilities are equal (\( p = q = 1/2 \)), the component term \( p^r q^{n-r} \) simplifies beautifully to just \( \frac{1}{2^n} \), regardless of the value of \( r \). This lets you compute the final fraction much faster!

Answer 1

The Correct Option is A

Step 1: Write the two facts.
For a binomial distribution, mean $= np = 4$ and variance $= npq = 2$.
Step 2: Find q.
Divide variance by mean: $\dfrac{npq}{np} = \dfrac{2}{4}$, so $q = \tfrac{1}{2}$.
Step 3: Find p.
Since $p+q=1$, $p = \tfrac{1}{2}$.
Step 4: Find n.
From $np = 4$ with $p=\tfrac{1}{2}$, we get $n = 8$.
Step 5: Use the binomial formula.
$P(X=2) = \binom{8}{2}\left(\tfrac{1}{2}\right)^2\left(\tfrac{1}{2}\right)^6 = \binom{8}{2}\left(\tfrac{1}{2}\right)^8$. Here $\binom{8}{2} = 28$ and $\left(\tfrac{1}{2}\right)^8 = \tfrac{1}{256}$.
Step 6: Simplify.
\[ P(X=2) = \frac{28}{256} = \frac{7}{64} \] \[ \boxed{\tfrac{7}{64}} \]