The random variable X follows binomial distribution B(n, p), for which the difference of the mean and the variance is 1 . If 2 P(X=2)=3 P(X=1), then n² P(X>1) is equal to :

Solution

Maths sir · 5 steps

Turn both conditions into equations in n and p, solve them together, then use the complement for P(X > 1).

1. Mean minus variance. Mean is np and variance is npq with q = 1 - p. So np - npq = np(1 - q) = np² = 1.
2. Probability condition. Write 2C(n, 2)p²qⁿ⁻² = 3C(n, 1)pqⁿ⁻¹, which is n(n-1)p²qⁿ⁻² = 3npqⁿ⁻¹. Cancel npqⁿ⁻² to get (n-1)p = 3q = 3 - 3p, so (n+2)p = 3.

   n(3/(n+2))² = 1 ⇒ n² - 5n + 4 = 0 ⇒ n = 1 or 4

3. Pick n. With n = 1, X = 2 is impossible, so n = 4 and p = 3/6 = 1/2. Check: np² = 4 × 1/4 = 1.
4. Use the complement. P(X > 1) = 1 - P(X = 0) - P(X = 1) = 1 - 1/16 - 4/16 = 11/16.
5. So. n² P(X > 1) = 16 × 11/16 = 11, which is option (c).

The idea to keep. In a binomial question, mean minus variance is np². A ratio of two neighbouring probabilities cancels most factors and leaves a simple linear equation in n and p.

Answer from the NTA final answer key. Solution by PrepWiser.