Mean and Variance of a binomial distribution are 24 and 128

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 :

$\frac{{33}}{{{2^{32}}}}$
$\frac{{33}}{{{2^{29}}}}$
$\frac{{33}}{{{2^{28}}}}$
$\frac{{33}}{{{2^{27}}}}$

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

$np = \frac{{24}}{{1 + q}}$ ……………….(2)

Product of mean and variance of binomial distribution = 128

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

(np) (npq) = 128

${\left( {np} \right)^2}q = 128$

Using Equation (2)

$\begin{gathered} \Rightarrow {\left( {\frac{{24}}{{1 + q}}} \right)^2}q = 128 \hfill \\ \Rightarrow \frac{{576q}}{{{{(1 + q)}^2}}} = 128 \hfill \\ \Rightarrow 576q = 128{(1 + q)^2} \hfill \\ \Rightarrow 576q = 128(1 + {q^2} + 2q) \hfill \\ \Rightarrow 576q = 128 + 128{q^2} + 256q \hfill \\ \Rightarrow 128{q^2} + 256q - 576q + 128 = 0 \hfill \\ \Rightarrow 128{q^2} - 320q + 128 = 0 \hfill \\ \Rightarrow 64(2{q^2} - 5q + 2) = 0 \hfill \\ \Rightarrow 2{q^2} - 5q + 2 = 0 \hfill \\ \end{gathered}$

On factorisation, we get

$\Rightarrow \left( {q - 2} \right)\left( {2q - 1} \right) = 0$

$\Rightarrow q = 2$ \text{ or } $q = \frac{1}{2}$

When $q = 2$ $\Rightarrow$ p = 1 – q

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

When $q = \frac{1}{2}$ $\Rightarrow$ p = 1 – q

$\Rightarrow p = 1 - \frac{1}{2} = \frac{1}{2}$

So, we get p = q = $\frac{1}{2}$

Using equation (2), n = 32

Step 3 of 3:

The probability for binomial distribution is:

$p(X = x) = {}^n{C_x}{p^x}{q^{n - x}}$

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

= ${}^{32}{C_1}{p^1}{q^{32 - 1}}$ + ${}^{32}{C_2}{p^2}{q^{32 - 2}}$

= ${}^{32}{C_1}\left( {\frac{1}{2}} \right){\left( {\frac{1}{2}} \right)^{31}}$ + ${}^{32}{C_1}{\left( {\frac{1}{2}} \right)^2}{\left( {\frac{1}{2}} \right)^{30}}$

= $\left( {32 + \frac{{32 \times 31}}{2}} \right)\left( {\frac{1}{{{2^{32}}}}} \right)$

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

Final Answer:

Hence, Option (C) is correct.

Solution:

Step 1 of 3:

Step 2 of 3:

Step 3 of 3:

Final Answer:

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