Question

If \( X \sim B(6, p) \) is a binomial variate and \[ \frac{P(X=4)}{P(X=2)} = \frac{1}{9}, \] then the value of \( p \) is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{4} \)
  \

Hint:

For binomial probability problems involving ratios, use the probability mass function and simplify using binomial coefficients. Solve for \( p \) algebraically.

Answer 1

The Correct Option is D

Step 1: Define the Binomial Probability Formula
For a binomial distribution \( X \sim B(n, p) \), the probability mass function is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \] Given \( X \sim B(6, p) \), we apply this formula to compute \( P(X=4) \) and \( P(X=2) \). Step 2: Compute Probability Ratios
Using the binomial formula: \[ P(X=4) = \binom{6}{4} p^4 (1 - p)^2 \] \[ P(X=2) = \binom{6}{2} p^2 (1 - p)^4 \] Taking their ratio: \[ \frac{P(X=4)}{P(X=2)} = \frac{\binom{6}{4} p^4 (1 - p)^2}{\binom{6}{2} p^2 (1 - p)^4} \] Substituting \( \binom{6}{4} = \binom{6}{2} = 15 \): \[ \frac{15 p^4 (1 - p)^2}{15 p^2 (1 - p)^4} \] \[ = \frac{p^4}{p^2} \times \frac{(1 - p)^2}{(1 - p)^4} \] \[ = p^2 \times \frac{1}{(1 - p)^2} \] Since we are given: \[ \frac{P(X=4)}{P(X=2)} = \frac{1}{9} \] we equate: \[ p^2 \times \frac{1}{(1 - p)^2} = \frac{1}{9} \] Step 3: Solve for \( p \)
Taking the square root on both sides: \[ \frac{p}{1 - p} = \frac{1}{3} \] \[ 3p = 1 - p \] \[ 4p = 1 \] \[ p = \frac{1}{4} \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{\frac{1}{4}} \] \bigskip