Question

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 :

• A. $\frac{33}{2^{32}}$
• B. $\frac{33}{2^{29}}$
• C. $\frac{33}{2^{28}}$
• D. $\frac{33}{2^{27}}$

Answer 1

Answer: (C)

$n p+n p q=24 ....$(1)
$n p \cdot n p q=128....$(2)
Solving (1) and (2):
We get $p =\frac{1}{2}, q =\frac{1}{2}, n =32$.
Now,
$P ( X =1)+ P ( X =2)$
$={ }^{32} C _1 pq ^{31}+{ }^{32} C _2 p ^2 q ^{30}$
$=\frac{33}{2^{28}}$