1) In a binomial distribution , mean is 18 and variance is 12 then p= A) $\frac{2}{3}$ B) $\frac{1}{3}$ C) $\frac{3}{4}$ D) $\frac{1}{2}$ Answer: Option BExplanation: We know that , mean =np=18 and variance = n p q =12 Now, $\frac{mpq}{np}=\frac{12}{18}\Rightarrow q=\frac{2}{3}$ $\therefore$ $p=1-q=1-\frac{2}{3}=\frac{1}{3}$
