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)

$\frac{33}{2^{28}}$

Answer 2

If n is the number of trials, p is the probability of success and q is a probability of unsuccess then,
Mean = np and variance = npq.
Here _np + npq = _24 …(i)|
np.npq = 128 …(ii)
and q = 1 – p …(iii)
from eq. (i), (ii) and (iii) :
p=q=$\frac{1}{2}$
and n = 32.
∴ Required probability
=P(X=1)+P(X=2)
=32C1⋅($\frac{1}{2}$)32+32C2⋅($\frac{1}{2}$)32
=(32+$\frac{32\times31}{2}$)⋅$\frac{1}{2}$32
=$\frac{33}{2^{28}}$