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

## Solution:

Tip for solving this question:

Firstly find the mean and variance of the binomial distribution.

Make an equation according to the question and solve it.

Then use the formula of probability for the binomial distribution.

### Step 1 of 3:

The mean of the binomial distribution is:

Mean = np

And the variance of the binomial distribution is:

Variance = npq

Where n is the number of trials

P is the probability of success

q is probability of failure i.e. q = 1- p ……….(1)

### Step 2 of 3:

According to the question,

Sum of mean and variance of binomial distribution = 24

i.e. mean + variance = 24

np + npq = 24

np(1 + q) = 24

……………….(2)

Product of mean and variance of binomial distribution = 128

i.e. (mean)(product) = 128

(np) (npq) = 128

Using Equation (2)

On factorisation, we get

\text{ or }

When p = 1 – q

p = 1 – 2 = -1 ………..Not considered due to negative

When p = 1 – q

So, we get p = q =

Using equation (2), n = 32

### Step 3 of 3:

The probability for binomial distribution is:

probability of one or two successes = p(X = 1) + p(X = 2)

= +

= +

=

=

## Final Answer:

Hence, Option (C) is correct.

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