If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :

Solution

Maths sir · 4 steps

Mean and variance are the two roots of one quadratic; mean is the bigger one, and that gives n and p.

1. Set up a quadratic. Mean is np and variance is npq. Their sum is 24 and their product is 128, so both are roots of t² - 24t + 128 = 0. That gives t = 16 or t = 8.
2. Find n and p. Variance is always smaller than the mean because q < 1, so np = 16 and npq = 8. Then q = 8/16 = 1/2, so p = 1/2 and n = 32.

   P(X=1)+P(X=2) = (C(32, 1)+C(32, 2))1/2³² = (32+496)/2³² = 528/2³²

3. Cancel powers of 2. 528 = 16 × 33 = 2⁴ × 33. So the probability is (2⁴ × 33)/2³² = 33/2²⁸.
4. So. The probability of one or two successes is 33/2²⁸, which is option (c).

The idea to keep. When a question gives you the sum and the product of mean and variance, make a quadratic. The larger root is the mean, because variance = np × q and q < 1.

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