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

MCQ

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

A
$2 / 3$
B
$1 / 3$
C
$3 / 4$
D
$1 / 2$

✓

Answer

(b) : Given, mean, $E(X)=18$
And $\operatorname{Var}(X)=12$
But $E(X)=n p$ and $\operatorname{Var}(X)=n p q$
$\therefore n p=18$ and $n p q=12$
Now, $\frac{n p q}{n p}=\frac{12}{18} \Rightarrow q=\frac{2}{3} \quad \therefore p=1-q=1-\frac{2}{3}=\frac{1}{3}$

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