Question

If \( X \sim B(n, p) \) and \( E(X) = 6 \) and \( \text{Var}(X) = 4.2 \), then find \( n \) and \( p \).

Hint:

For binomial distribution, use \( E(X) = np \), \( \text{Var}(X) = np(1 - p) \); solve simultaneously for \( n \) and \( p \).

Answer 1

For binomial distribution \( X \sim B(n, p) \): 
\[ E(X) = np = 6, \text{Var}(X) = np(1 - p) = 4.2. \] From \( np = 6 \), we have \( n = \frac{6}{p} \). 
Substitute into variance: 
\[ np(1 - p) = 6 (1 - p) = 4.2 \Rightarrow 1 - p = \frac{4.2}{6} = 0.7 \Rightarrow p = 0.3. \] \[ n = \frac{6}{p} = \frac{6}{0.3} = 20. \] Verify: \( \text{Var}(X) = 20 \cdot 0.3 \cdot 0.7 = 20 \cdot 0.21 = 4.2 \), correct. 
Answer: \( n = 20 \), \( p = 0.3 \).