Question

The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P (X = 0), P (X = 1) and P (X ≥ 2).

 

Answer 1

Given: mean =16 and variance = 8

Let n and p be the parameters of the distribution.
 

That is, np = 16 and npq  = 8

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

\[\text{ and } p = 1 - q = \frac{1}{2}\]

\[ np = 16 \]

\[ \Rightarrow n = 32\]

\[ \therefore P(X = r) = ^{32}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{32 - r} , r = 0, 1, 2 . . . . 32, \]

\[ \Rightarrow P(X = 0) = \left( \frac{1}{2} \right)^{32} \]

\[ P(X = 1) = 32 \left( \frac{1}{2} \right)^{32} = \left( \frac{1}{2} \right)^2 \]

\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1)\]

\[ = 1 - \left( \frac{1}{2} \right)^{32} - \left( \frac{1}{2} \right)^{27} \]

\[ = 1 - \left( \frac{1 + 32}{2^{32}} \right) \]

\[ = 1 - \frac{33}{2^{32}}\]