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Q.
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 :
$n p+n p q=24 ....$(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} $
$=\frac{33}{2^{28}}$