Test Index

Probability Part 2

Section: Mathematics

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Question : 71 of 100

Marks: +1, -0

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

[25-Jul-2022-Shift-1]

$\;\frac{33}{2^{32}}$
$\;\frac{33}{2^{29}}$
$\;\frac{33}{2^{28}}$
$\;\frac{33}{2^{27}}$

Solution: 👈: Video Solution

If

n

is number of trails,

p

is probability of success and

q

is probability of unsuccess then,

Mean

=n p

and variance

=n p q

.

Here

n p+n p q=24 \;\; \ldots(i)

n p . n p q=128 \;\; \ldots

(ii)

and

q=1-p \;\; \ldots(iii)

from eq. (i), (ii) and (iii) :

p=q=\;\frac{1}{2}

and

n=32

.

\therefore

Required probability

=p(X=1)+p(X=2)

= {}^{32}C_1\cdot\left(\frac{1}{2}\right)^{32} + {}^{32}C_2\cdot\left(\frac{1}{2}\right)^{32}

= \left(32+\;\frac{32 \times 31}{2}\right) \cdot\;\frac{1}{2^{32}}

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

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