Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If P(X>n-3)=k/2ⁿ, then k is equal to :

Solution

Maths sir · 4 steps

Mean and variance are roots of one quadratic, and the mean is the bigger root. That fixes n and p.

1. Set up a quadratic. Mean np and variance npq have sum 24 and product 128. So they are the roots of t² - 24t + 128 = 0, which gives t = 16 and t = 8.
2. Find n and p. Variance npq is mean times q, and q < 1, so the mean is the larger root: np = 16, npq = 8. Dividing gives q = 1/2, so p = 1/2 and n = 32.
3. Read the inequality. X > n - 3 means X > 29, so X is 30, 31 or 32. With p = q = 1/2, every outcome carries the same factor 1/2³².

   P(X > 29) = (C(32, 30) + C(32, 31) + C(32, 32))/2³² = (496 + 32 + 1)/2³²

4. So. P(X > 29) = 529/2³², so k = 529, which is option (b).

The idea to keep. When you know the sum and product of mean and variance, put them in a quadratic. The mean is always the larger root, because variance = mean × q.

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