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UPSC NDA I 2015 Mathematics Paper

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Question : 115 of 120

Marks: +1, -0

The mean and the variance in a binomial distribution are found to be 2 and 1 respectively. The probability P(

) is

$1/2$
$1/4$
$1/8$
$1/16$

Solution:

Concept:

Mean =

np

, Variance =

npq

for a binomial distribution.

Explanation:

Given: Mean

np = 2

, Variance

npq = 1

.

Divide variance by mean:

\frac{npq}{np} = q = \frac{1}{2}

.

Since

p + q = 1

, we get

p = 1 - \frac{1}{2} = \frac{1}{2}

.

From

np = 2

, substitute

p = \frac{1}{2}

:

n \cdot \frac{1}{2} = 2

so

n = 4

.

Probability

P(X = 0) = \binom{4}{0} p^0 q^{4} = 1 \cdot 1 \cdot \left(\frac{1}{2}\right)^4 = \frac{1}{16}

.

Answer:

\frac{1}{16}

(Option D).

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