If the sum and product of mean and variance of a binomial distribution are 24 and 128 resp…

Question

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

Accepted Answer

$np+npq=24$, $np\cdot npq=128$. Solving gives $p=q=\dfrac12$, $n=32$.

$$P(1)+P(2) = {}^{32}C_1\left(\frac12\right)^{32}+{}^{32}C_2\left(\frac12\right)^{32} = \frac{32+496}{2^{32}}=\frac{33}{2^{28}}$$

Solutions · 1