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The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is:

  • Option 1)

    128/256

  • Option 2)

    219/256

  • Option 3)

    37/256

  • Option 4)

    28/256

 

Answers (1)

best_answer

As we learnt in 

Binomial Distribution(Statistical) -

Mean = np

Variance = npq

Standard \: deviation =\sqrt{npq}

-

 

 mean = u_{x}= 4

variance = a^{2}x = 2

Binomial Distribution,

\mu _{x} = np = 4 .................. (1)

\sigma _{x^2} = np \left ( 1-p \right ) =2 ............(2)

solving \left ( 1 \right ) we get\ 1-p=\frac{1}{2}

\Rightarrow p=0.5

Similarly, n = 8

Thus, Probability of 2 Sucess i.e p\left ( x=2 \right )

=\: ^{n}c_{x}\: p^{x} \left ( 1-p \right )^{n-x}

=\ ^{8}_{2}C \left ( 0.5 \right )^2\left ( 0.5 \right )^6

= \frac{8\times 7}{2}\: \left ( 0.5 \right )^8

= \frac{56}{2}\:\times \frac{1}{2^{8}}

=\:\frac{28}{256}


Option 1)

128/256

Incorrect

Option 2)

219/256

Incorrect

Option 3)

37/256

Incorrect

Option 4)

28/256

Correct

Posted by

divya.saini

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