Question

In a Binomial distribution, if \(n\) is the number of trials and the mean and variance are \(4\) and \(3\) respectively, then \[ 2^{32}\cdot P\left(X=\frac{n}{2}\right)= \]

• A. \({}^{16}C_8(3^8)\)
• B. \({}^{12}C_6(2^6)\)
• C. \({}^{32}C_{16}(3^{16})\)
• D. \({}^{16}C_7(3^9)\)

Hint:

For Binomial distribution: \[ \text{Mean}=np, \qquad \text{Variance}=npq. \] Use these first to determine \(n\), \(p\), and \(q\), then apply the probability formula: \[ P(X=r)={}^{n}C_r p^r q^{n-r}. \]

Answer 1

The Correct Option is A

Step 1: Identify the binomial distribution parameters.
For a Binomial distribution $B(n, p)$: mean $= np = 4$ and variance $= np(1-p) = 3$.
Step 2: Solve for $p$ and $n$.
$\frac{\text{variance}}{\text{mean}} = 1-p = \frac{3}{4}$, so $p = \frac{1}{4}$ and $q = 1-p = \frac{3}{4}$. Then $n = \frac{4}{p} = \frac{4}{1/4} = 16$.
Step 3: Find the value of $n/2$.
$\frac{n}{2} = \frac{16}{2} = 8$. So we need $P(X = 8)$.
Step 4: Write $P(X = 8)$ using the binomial formula.
$P(X=8) = \binom{16}{8} p^8 q^8 = \binom{16}{8} \left(\frac{1}{4}\right)^8 \left(\frac{3}{4}\right)^8 = \binom{16}{8} \frac{3^8}{4^{16}}$.
Step 5: Compute $2^{32} \cdot P(X=8)$.
$2^{32} \cdot P(X=8) = 2^{32} \cdot \binom{16}{8} \frac{3^8}{4^{16}} = 2^{32} \cdot \binom{16}{8} \frac{3^8}{2^{32}} = \binom{16}{8} \cdot 3^8$.
Since $\binom{16}{8} = {}^{16}C_8$, we get ${}^{16}C_8 \cdot 3^8$.
Step 6: Match with options.
${}^{16}C_8 \cdot 3^8$ corresponds to option (1): ${}^{16}C_8(3^8)$.
\[ \boxed{{}^{16}C_8 \cdot 3^8} \]