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  • Need clarity, kindly explain! - Statistics and Probability - JEE Main

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X=1) is

  • Option 1)

    1/16

  • Option 2)

    1/8

  • Option 3)

    1/4

  • Option 4)

    1/32

 

Answers (2)

As we learnt in 

Binomial Distribution(Statistical) -

Mean = np

Variance = npq

Standard \: deviation =\sqrt{npq}

-

 

 Mean=np=4 ........ (1)

Variance=np(1-p)=2 .............. (2)

From (1) and (2), \frac{np(1-p)}{np}=\frac{1}{2}

                          \Rightarrow P=0.5

From (1), n=8

Thus, P(X=1)= _{}^{8}\textrm{C}_{1}P^{1}(1-P)^{8-1}

                   = 8\times (0.5)^{8}

                    = 8\times \frac{1}{2^{8}}

                    =\frac{1}{2^{5}}

                   = \frac{1}{32}


Option 1)

1/16

This option is incorrect

Option 2)

1/8

This option is incorrect

Option 3)

1/4

This option is incorrect

Option 4)

1/32

This option is correct

Posted by

Sabhrant Ambastha

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JEE Main high-scoring chapters and topics

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As it is give in the question that:

mean(np)=4 (equation.....1)

variance(npq)=2(equation....2)

now divided equ(2)/equ(1);then:

npq/np=2/4

q=1/2

now as we know that sum of the probability=1

p+q=1

p+1/2=1

p=1-1/2

p=1/2

now substitute p value in equation (1)

np=4

n(1/2)=4

n=4*2

n=8

now as we know that p(X=1)

by the general formula p(X=R)=ncr (p)^r (q)^n-r

now from th general formula above:

p(X=1)=8c1(1/2)^ (1/2)^8-1

           =8*(1/2)^7+1

           =8*(1/2)^8

           =8*1/256

p(X=R) =1/32

therefore this is the whole procedure for the above question

Posted by

kinshuk thatikonda

View full answer

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