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:
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For binomial distributions, use the formulas for mean \((np)\) and variance \((np(1-p))\) directly, and carefully simplify conditions involving probabilities.
Given a Binomial Distribution \( B(7, p) \) with \( n = 7 \) and \( p = p \). We are provided that \( P(x = 3) = 5P(x = 4) \). Using the binomial probability formula \( P(x = k) = \binom{n}{k} p^k (1 - p)^{n-k} \), and substituting \( n = 7 \), we get \( \binom{7}{3} p^3 (1 - p)^4 = 5 \cdot \binom{7}{4} p^4 (1 - p)^3 \). Simplifying the ratio of binomial coefficients gives \( \frac{\binom{7}{3}}{\binom{7}{4}} = \frac{p}{1 - p} \). Calculating the coefficients: \( \binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} \) and \( \binom{7}{4} = \frac{7 \cdot 6 \cdot 5 \cdot 4}{4 \cdot 3 \cdot 2 \cdot 1} \). Further simplification leads to \( 1 - p = 5p \). Solving for \( p \): \( 6p = 1 \implies p = \frac{1}{6} \). Consequently, \( q = 1 - p = \frac{5}{6} \). Substituting \( n = 7, p = \frac{1}{6}, q = \frac{5}{6} \), the mean is \( np = 7 \times \frac{1}{6} = \frac{7}{6} \) and the variance is \( npq = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36} \). The sum of the mean and variance is \( \frac{7}{6} + \frac{35}{36} \). Simplifying the fractions, \( \frac{7}{6} = \frac{42}{36} \), so the sum is \( \frac{42}{36} + \frac{35}{36} = \frac{77}{36} \). Final Answer: \( \frac{77}{36} \).