The mean and variance of a random variable X having a binomial distri…

The correct answer is Option (4) → A-IV, B-I, C-II, D-III

For a binomial distribution, mean = np and variance = npq.

    \( \displaystyle np=1,\ npq=\frac{2}{3} \)

    \( \displaystyle q=\frac{npq}{np}=\frac{2}{3} \)

    \( \displaystyle p=1-q=\frac{1}{3},\quad n=\frac{1}{p}=3 \)

So \( P(X=r)={}^{3}C_{r}\left(\frac{1}{3}\right)^r\left(\frac{2}{3}\right)^{3-r} \).

(A) \( P(X=0) \)

This matches (IV) 8/27.

    \( \displaystyle P(X=0)=\left(\frac{2}{3}\right)^3=\frac{8}{27} \)

(B) \( P(X=3) \)

This matches (I) 1/27.

    \( \displaystyle P(X=3)=\left(\frac{1}{3}\right)^3=\frac{1}{27} \)

(C) \( P(X=2) \)

This matches (II) 2/9.

    \( \displaystyle P(X=2)=3\cdot\frac{1}{9}\cdot\frac{2}{3}=\frac{6}{27} \)

    \( \displaystyle =\frac{2}{9} \)

(D) \( P(X=1) \)

This matches (III) 4/9.

    \( \displaystyle P(X=1)=3\cdot\frac{1}{3}\cdot\frac{4}{9}=\frac{12}{27} \)

    \( \displaystyle =\frac{4}{9} \)

So the matching is A-IV, B-I, C-II, D-III.