Question

Let \(X\) be a random variable having binomial distribution \(B(7, p)\). If \(P(X = 3) = 5P(X = 4)\), then the sum of the mean and the variance of \(X\) is:

• A. \(\frac{105}{16}\)
• B. \(\frac{7}{16}\)
• C. \(\frac{77}{36}\)
• D. \(\frac{49}{16}\)

Hint:

For binomial distributions, use the formulas for mean \((np)\) and variance \((np(1-p))\) directly, and carefully simplify conditions involving probabilities.

Answer 1

The Correct Option is C

Given a Binomial Distribution: \( B(7, p) \), where \[ n = 7, \quad p = p \] We are provided: \[ P(x = 3) = 5P(x = 4) \] Using the binomial probability formula: \[ P(x = k) = \binom{n}{k} p^k (1 - p)^{n-k} \] Substitute \( n = 7 \) into the formula: \[ \binom{7}{3} p^3 (1 - p)^4 = 5 \cdot \binom{7}{4} p^4 (1 - p)^3 \] Simplify the binomial coefficients: \[ \frac{\binom{7}{3}}{\binom{7}{4}} = \frac{p}{1 - p} \] \[ \binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}, \quad \binom{7}{4} = \frac{7 \cdot 6 \cdot 5 \cdot 4}{4 \cdot 3 \cdot 2 \cdot 1} \] Simplifying further: \[ \frac{\binom{7}{3}}{\binom{7}{4}} = \frac{p}{1 - p} \] Equating the powers of \( p \) and \( (1-p) \): \[ 1 - p = 5p \] Solve for \( p \): \[ 6p = 1 \implies p = \frac{1}{6} \] Thus: \[ q = 1 - p = \frac{5}{6} \] Substitute \( n = 7, p = \frac{1}{6}, q = \frac{5}{6} \): \[ \text{Mean} = np = 7 \times \frac{1}{6} = \frac{7}{6} \] \[ \text{Variance} = npq = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36} \] Finally, the sum of the mean and variance: \[ \text{Sum} = \frac{7}{6} + \frac{35}{36} \] Simplify the fractions: \[ \frac{7}{6} = \frac{42}{36}, \quad \text{so, } \] \[ \text{Sum} = \frac{42}{36} + \frac{35}{36} = \frac{77}{36} \] Final Answer: \( \frac{77}{36} \).