Question

If a random variable $X$ follows the binomial distribution with parameters $n = 5$, $p$, and $P(X = 2) = 9P(X = 3)$, then $p$ is equal to:

• A. $10$
• B. $\frac{1}{10}$
• C. $5$
• D. $1\frac{1}{5}$

Answer 1

The Correct Option is B

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) \( \frac{1}{10} \)

Answer 2

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: \[ 1 = 9p + p \implies 1 = 10p \implies p = \frac{1}{10}. \]