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. \(\frac{49}{16}\)

Answer 1

The Correct Option is C

1. Given that the random variable \(X\) follows a binomial distribution \(B(7, p)\), it means \(X\) can take values from 0 to 7 with successes occurring with probability \(p\). The probability mass function for a binomial distribution is given by:
   P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
   where \(n\) is the number of trials and \(k\) is the number of successful trials or 'x'. Here, \(n = 7\).
2. According to the question, \(P(X = 3) = 5P(X = 4)\). Using the binomial formula:
   P(X = 3) = \binom{7}{3} p^3 (1-p)^{4}
   P(X = 4) = \binom{7}{4} p^4 (1-p)^{3}
3. Setting up the equation provided in the problem:
   \binom{7}{3} p^3 (1-p)^4 = 5 \cdot \binom{7}{4} p^4 (1-p)^3
   \frac{\binom{7}{3}}{\binom{7}{4}} \cdot \frac{(1-p)}{p} = 5
   After calculating the binomial coefficients:
   \frac{35}{35} \cdot \frac{(1-p)}{p} = 5
   (1-p) = 5p
   1 = 6p
   p = \frac{1}{6}
4. Now, the mean and variance of a binomial distribution can be calculated by:
   \text{Mean} = np = 7 \times \frac{1}{6} = \frac{7}{6}
   \text{Variance} = np(1-p) = 7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}
5. The sum of the mean and variance is:
   \frac{7}{6} + \frac{35}{36} = \frac{42}{36} + \frac{35}{36} = \frac{77}{36}
6. Therefore, the sum of the mean and the variance is \(\frac{77}{36}\), which matches the given correct answer.