Question

The Variance of the following probability distribution of X is

X	\(0\)	\(1\)	\(2\)	\(3\)
P(X)	\(q^3\)	\(3q^2p\)	\(3qp^2\)	\(p^3\)

where \(0 < p < 1,q = 1-p\)

• A. \(3p^2q\)
• B. \(3pq\)
• C. \(3pq^2\)
• D. \(3q\)

Hint:

This is a binomial distribution with n = 3.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The probabilities \(q^3,\ 3q^2p,\ 3qp^2,\ p^3\) are those of a binomial distribution with \(n=3\) and success probability \(p\).

Step 2: Recall:
For a binomial distribution the mean is \(np\) and the variance is \(npq\).

Step 3: Apply:
With \(n=3\): variance \(=3pq\).

Step 4: Verify directly:
Mean \(=\sum xP=3q^2p+6qp^2+3p^3=3p(q^2+2qp+p^2)=3p(q+p)^2=3p\). And \(E(X^2)=3q^2p+12qp^2+9p^3=3p(q^2+4qp+3p^2)=3p(q+p)(q+3p)=3p(1+2p)\). So variance \(=3p+6p^2-9p^2=3p-3p^2=3p(1-p)=3pq\).

Step 5: Choose:
Option (B).

Final Answer:
The variance is 3pq. \[ \boxed{3pq} \]