1. Understand the binomial distribution:
Given \( X \sim \text{Binomial}(n=5, p) \) and \( P(X=2) = 9P(X=3) \).
2. Write the binomial probabilities:
\[ P(X=2) = \binom{5}{2} p^2 (1-p)^3 = 10 p^2 (1-p)^3 \] \[ P(X=3) = \binom{5}{3} p^3 (1-p)^2 = 10 p^3 (1-p)^2 \]
3. Set up the equation:
Given \( P(X=2) = 9P(X=3) \):
\[ 10 p^2 (1-p)^3 = 9 \cdot 10 p^3 (1-p)^2 \]
Simplify:
\[ (1-p) = 9p \implies 1 = 10p \implies p = \frac{1}{10} \]
Correct Answer: (B) \( \frac{1}{10} \)
The probability mass function of a binomial distribution is: \[ P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r}. \] Given $P(X = 2) = 9P(X = 3)$, substitute into the formula: \[ \binom{5}{2} p^2 (1 - p)^3 = 9 \binom{5}{3} p^3 (1 - p)^2. \] Simplify the binomial coefficients: \[ \frac{5 \cdot 4}{2} \cdot p^2 (1 - p)^3 = 9 \cdot \frac{5 \cdot 4 \cdot 3}{6} \cdot p^3 (1 - p)^2. \] \[ 10p^2(1 - p)^3 = 90p^3(1 - p)^2. \] Divide both sides by $10p^2(1 - p)^2$: \[ 1 - p = 9p. \] Simplify: \[ 1 = 9p + p \implies 1 = 10p \implies p = \frac{1}{10}. \]
\[ \begin{array}{c|c} X = x & P(X = x) \\ \hline 1 & 3K^2 \\ 3 & K \\ 5 & K^2 \\ 2 & 2K \end{array} \]