Question

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X=1) is......,Select Proper option from the given options.
(a) 1/16
(b) 1/8
(c) 1/4
(d) 1/32

Answer 1

Answer:

1/32

Step-by-step explanation:

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively

np = 4

npq = 2

npq/np = 2/4

=> q = 1/2

=> p = 1 - 1/2 = 1/2

n (1/2) = 4

=> n = 8

P(1) = ⁿC₁ p¹ q⁷  =  ⁸C₁ (1/2)¹ (1/2)⁷  =  8/ 256

= 1/32

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If a trial is repeated n times and p is the probability of Success then By BINOMIAL DISTRIBUTION the probability of R successes is

$\sf{ P( x = r) = \large{ {}^{n} C_r} \: {p}^{r} {(1 - p)}^{n - r}}$

Mean = np

Variance = np(1-p)

GIVEN

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively

TO CHOOSE THE CORRECT OPTION

P(X=1) =

(a) 1/16

(b) 1/8

(c) 1/4

(d) 1/32

CALCULATION

By the given condition

$\sf{np = 4} \: \: and \: \: np(1 - p) = 2$

So

$\displaystyle \sf{ 1 - p = \frac{2}{4} \: \: }$

$\implies \: \displaystyle \sf{ 1 - p = \frac{1}{2} \: \: }$

$\implies \: \displaystyle \sf{p = \frac{1}{2} \: \: }$

Again

$\displaystyle \sf{ np = 4\: \: } \: \: gives$

$\implies \: \displaystyle \sf{n = \frac{4}{p} = 8\: \: }$

Hence

$\sf{ P( x = 1) = \large{ {}^{n} C_1} \: {p}^{1} {(1 - p)}^{n - 1}}$

$\displaystyle \implies \: \sf{ P( x = 1) = \large{ {}^{8} C_1} \: \times \frac{1}{2} \times { \bigg(1 - \frac{1}{2} \bigg)}^{8 - 1}}$

$\displaystyle \implies \: \sf{ P( x = 1) = 8 \: \times { \bigg( \frac{1}{2} \bigg)}^{8 }}$

$\displaystyle \implies \: \sf{ P( x = 1) = \: { \bigg( \frac{1}{2} \bigg)}^{5 }}$

$\displaystyle \implies \: \sf{ P( x = 1) = \: \frac{1}{32} }$

RESULT

$\boxed{ \displaystyle\: \sf{ P( x = 1) = \: \frac{1}{32} } \: \: \: }$

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