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# If the Mean and Variance of a Binomial Distribution Are 4 and 3, Respectively, Find the Probability of No Success. - Mathematics

ConceptBernoulli Trials and Binomial Distribution

#### Question

If the mean and variance of a binomial distribution are 4 and 3, respectively, find the probability of no success.

#### Solution

Mean (np)= 4
Variance (npq )= 3

$\Rightarrow q = \frac{3}{4}$
$\text{ Hence, } p = 1 - \frac{3}{4} = \frac{1}{4}$
$\text{ and } n = \frac{\text{ Mean }}{p} = 4 \times 4 = 16$
$\text{ Therefore, the binomial distribution is given by }$
$P(X = r) =^ {16}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{16 - r} , r = 0, 1, 2 . . . . r$
$\text{ Probability of no success } = ^{16}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{16 - 0} = \left( \frac{3}{4} \right)^{16}$

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Solution If the Mean and Variance of a Binomial Distribution Are 4 and 3, Respectively, Find the Probability of No Success. Concept: Bernoulli Trials and Binomial Distribution.
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