Mathematics | MHT CET 2019 | Discussion Page

In a binomial distribution , mean is 18 and variance is 12 then p=

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{3}{4}$

$\frac{1}{2}$

Option B

We know that , mean =np=18

and variance = n p q =12

Now, $\frac{mpq}{np}=\frac{12}{18}\Rightarrow q=\frac{2}{3}$

$\therefore$ $p=1-q=1-\frac{2}{3}=\frac{1}{3}$

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