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 :

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) (33/232) (B) (33/229) (C) (33/228) (D) (33/227)

Tardigrade Admin · Accepted Answer

n p+n p q=24 ....(1) n p ⋅ n p q=128....(2) Solving (1) and (2): We get p =(1/2), q =(1/2), n =32. Now, P ( X =1)+ P ( X =2) = 32 C 1 pq 31+ 32 C 2 p 2 q 30 =(33/228)

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 :

$\frac{33}{2 ^{32}}$

$\frac{33}{2 ^{29}}$

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

$\frac{33}{2 ^{27}}$

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

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 :

23233​

22933​

22833​

22733​

np+npq=24....(1) np⋅npq=128....(2) Solving (1) and (2): We get p=21​,q=21​,n=32. Now, P(X=1)+P(X=2) =32C1​pq31+32C2​p2q30 =22833​

$\frac{33}{2 ^{32}}$

$\frac{33}{2 ^{29}}$

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

$\frac{33}{2 ^{27}}$

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