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  • If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, the the probability of one or two successes is :Option: 1

If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, the the probability of one or two successes is :

Option: 1

\frac{33}{2^{32}}


Option: 2

\frac{33}{2^{29}}


Option: 3

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


Option: 4

\frac{33}{2^{27}}


Answers (1)

\mathrm{n p+n p q=24} \\          ........(1)

\mathrm{n p \cdot n p q=128}          .........(2)

Solving (1) and (2)

we get   \mathrm{p=\frac{1}{2}, q=\frac{1}{2}, n=32}

Now,

\mathrm{P(x=1)+P(x=2)} \\

\mathrm{={ }^{32} c_{1}pq^{31}+{ }^{32} c_{2} p^{2} q^{30}} \\

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

Hence correct option is 3

Posted by

Ramraj Saini

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