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If the mean and the variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is :

  • Option 1)

    \frac{1}{16}

  • Option 2)

    \frac{9}{16}

  • Option 3)

    \frac{3}{4}

  • Option 4)

    \frac{15}{16}

Answers (1)

best_answer

As we have learnt,

 

Binomial Distribution(Statistical) -

Mean = np

Variance = npq

Standard \: deviation =\sqrt{npq}

-

 

 

Binomial Distribution -

In a series of n independent trials if the probability of success P in each trial is same, then the probapility of r success is

P(X=r)= \left\{\begin{matrix} \left ( \frac{n}{r} \right )\: q^{n-r}\cdot P^{r} & q = 1-P\\ \left ( \frac{n}{r} \right )\frac{1}{2^{n}} &P=\frac{1}{2} \end{matrix}\right.

- wherein

Where \sum is probability of failure.

 

 \\\lambda = 2 \\\sigma^2 = 1 \\P(X\geq 1) = 1 - P(X=0)

                        \\=1 - \binom{n}{0}p^oq^n \\= 1 - q^n \\=1-(1-p)^n

where, np= 2\;\&\;npq =1

So, q = \frac{1}{2} = p\;\&\; n = 4

\\P(X \geq 1) = 1 - (1-\frac{1}{2})^4

                       = 1 - \frac{1}{16} = \frac{15}{16}

 


Option 1)

\frac{1}{16}

Option 2)

\frac{9}{16}

Option 3)

\frac{3}{4}

Option 4)

\frac{15}{16}

Posted by

gaurav

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