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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

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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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