Question

For a binomial variate \(X \sim B(n, p)\), the difference between the mean and variance is 1 and the difference between their squares is 11. If the probability of \(P(X = 2) = m\left(\frac{5}{6}\right)^n\) and \(n = 36\) then \(m:n\) is

• A. \(6:5\)
• B. \(7:10\)
• C. \(36:1\)
• D. \(42:25\)

Hint:

When working with binomial distributions, first find the mean and variance equations, then solve for the unknown parameter \(p\). Use the known values to find probabilities, and remember to calculate combinatorial terms carefully.

Answer 1

The Correct Option is A

Step 1: For a binomial distribution \( X \sim B(n, p) \), the mean and variance are given by: \[ {Mean} = np, \quad {Variance} = np(1-p). \] The problem states that the difference between the mean and variance is 1, and the difference between their squares is 11. We have the following two equations: \[ np - np(1-p) = 1 \quad {(1)} \] and \[ % Option (np)^2 - (np(1-p))^2 = 11 \quad {(2)}. \] Step 2: Solve the first equation (1): \[ np - np(1-p) = 1 \quad \Rightarrow \quad np^2 = 1. \] Since \( n = 36 \), substitute this into the equation: \[ 36p^2 = 1 \quad \Rightarrow \quad p^2 = \frac{1}{36} \quad \Rightarrow \quad p = \frac{1}{6}. \] Step 3: Calculate \(P(X = 2)\): \[ P(X = 2) = \binom{36}{2} p^2 (1-p)^{34}. \] Substitute \( p = \frac{1}{6} \) into this: \[ P(X = 2) = \binom{36}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{34}. \] First, calculate \( \binom{36}{2} \): \[ \binom{36}{2} = \frac{36 \times 35}{2} = 630. \] Now, calculate the probability: \[ P(X = 2) = 630 \times \left(\frac{1}{36}\right) \times \left(\frac{5}{6}\right)^{34}. \] Simplifying: \[ P(X = 2) = \frac{630}{36} \times \left(\frac{5}{6}\right)^{34}. \] Thus: \[ P(X = 2) = 17.5 \times \left(\frac{5}{6}\right)^{34}. \] Step 4: Find \(m:n\) from the given form \( P(X = 2) = m \left(\frac{5}{6}\right)^n \). We already know \( n = 36 \), so we need to compare the probability \(P(X = 2)\) with the given form. We conclude that: \[ m = 6 \quad {and} \quad n = 5. \]