Question

In a binomial distribution, if:
- Number of trials $ n $
- Mean $ = 4 $
- Variance $ = 3 $
Then find: $$ 2^{32} \cdot P\left(X = \frac{n}{2}\right) $$

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

Hint:

Use the identities: \( \mu = np \), \( \sigma^2 = npq \), and standard binomial formula to compute probability. Use exponent tricks to simplify.

Answer 1

The Correct Option is A

In a binomial distribution: - Mean \( \mu = np \) - Variance \( \sigma^2 = npq \) Given: \[ np = 4,\quad npq = 3 \] Divide the equations: \[ \frac{npq}{np} = \frac{3}{4} \Rightarrow q = \frac{3}{4} \Rightarrow p = \frac{1}{4} \Rightarrow np = 4 \Rightarrow n = \frac{4}{1/4} = 16 \] So: \[ n = 16,\quad p = \frac{1}{4},\quad q = \frac{3}{4} \] We are asked to find: \[ 2^{32} \cdot P(X = \frac{n}{2}) = 2^{32} \cdot P(X = 8) \] Binomial probability: \[ P(X = r) = ^nC_r \cdot p^r \cdot q^{n - r} = ^{16}C_8 \cdot \left(\frac{1}{4}\right)^8 \cdot \left(\frac{3}{4}\right)^8 = ^{16}C_8 \cdot \left( \frac{1 \cdot 3}{4^2} \right)^8 = ^{16}C_8 \cdot \left(\frac{3}{16}\right)^8 \] Now: \[ 2^{32} = (4^2)^8 = 16^8 \Rightarrow 2^{32} \cdot P(X = 8) = ^{16}C_8 \cdot \left(\frac{3}{16}\right)^8 \cdot 16^8 = ^{16}C_8 \cdot 3^8 \] \[ \Rightarrow \boxed{^{16}C_8 \cdot 3^8} \]