Question

A box contains 8 red and N green balls. Two balls are drawn at random from it. If X is the random variable representing the number of green balls drawn and \(E(X) = 1.2\), then \(N = \ldots\)

• A. \(4\)
• B. \(8\)
• C. \(12\)
• D. \(16\)

Hint:

\(E(X)\) for drawing 2 balls is \(2\times\frac{N}{N+8}\).

Answer 1

The Correct Option is C

Step 1: Use the hypergeometric distribution
$E(X)=\sum xP(x)=P(1)+2P(2)$ with $\binom{N+8}{2}$ outcomes.

Step 2: Solve
$E(X)=\dfrac{8N+2\binom N2}{\binom{N+8}2}=\dfrac{2N}{N+8}$. Setting it to 1.2 gives $N=12$, option (C).

Final Answer:
The number of green balls is 12, option (C). \[ \boxed{12} \]