Question

If \( X \sim B(7, p) \) is a binomial variate and \( P(X=3) = P(X=5) \) then \( p = \)

• A. \( \frac{5-\sqrt{10}}{3} \)
• B. \( \frac{\sqrt{10}-2}{3} \)
• C. \( \frac{5-\sqrt{15}}{2} \)
• D. \( \frac{\sqrt{15}-3}{2} \)

Hint:

Always check that the calculated probability value falls within the interval [0, 1].

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We are given a condition on a Binomial distribution probability mass function. We need to set up the equation using the formula and solve for the parameter \( p \).
Step 2: Key Formula or Approach:

For Binomial distribution \( B(n, p) \): \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Step 3: Detailed Explanation:

Given \( n=7 \) and \( P(X=3) = P(X=5) \). \[ \binom{7}{3} p^3 (1-p)^4 = \binom{7}{5} p^5 (1-p)^2 \] Substitute \( \binom{7}{3} = 35 \) and \( \binom{7}{5} = 21 \). Let \( q = 1-p \). \[ 35 p^3 q^4 = 21 p^5 q^2 \] Divide both sides by \( 7 p^3 q^2 \) (assuming \( p \neq 0, q \neq 0 \)): \[ 5 q^2 = 3 p^2 \] Substitute \( q = 1-p \): \[ 5 (1-p)^2 = 3 p^2 \] \[ 5 (1 - 2p + p^2) = 3 p^2 \] \[ 5 - 10p + 5p^2 = 3p^2 \] \[ 2p^2 - 10p + 5 = 0 \] Solve for \( p \) using the quadratic formula: \[ p = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(2)(5)}}{2(2)} \] \[ p = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm \sqrt{60}}{4} = \frac{10 \pm 2\sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2} \] Since \( p \) is a probability, \( 0 \le p \le 1 \). \( \sqrt{15} \approx 3.87 \). \( \frac{5 + 3.87}{2} \textgreater 1 \) (Rejected). \( \frac{5 - 3.87}{2} \textless 1 \) (Accepted). So, \( p = \frac{5 - \sqrt{15}}{2} \).
Step 4: Final Answer:

The value of \( p \) is \( \frac{5-\sqrt{15}}{2} \).