Question

The probability mass function (p.m.f.) of a random variable X is as follows:
\(P(X=0)=3k^{3},\ P(X=1)=4k-10k^{2},\ P(X=2)=5k-1,\ P(X=x)=0\) for any value of \(x\), then the value of \(k\) is:

• A. \(\frac{3}{4}\)
• B. \(\frac{3}{7}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{2}{3}\)

Hint:

The probabilities must add to 1 and each must be non-negative; solve the cubic and reject invalid roots.

Answer 1

The Correct Option is C

Step 1: Use total probability
\(3k^{3}+4k-10k^{2}+5k-1=1\), so \(3k^{3}-10k^{2}+9k-2=0\).

Step 2: Factorise
\((3k-1)(k-1)(k-2)=0\), so \(k=\frac{1}{3},1,2\).

Step 3: Check non-negativity
For \(k=1\): \(P(0)=3\), and for \(k=2\): \(P(0)=24\), which exceed 1. So these are rejected.

Step 4: Verify k = 1/3
\(P(0)=\frac19,\ P(1)=\frac29,\ P(2)=\frac23\), all non-negative and adding to 1.

Final Answer:
The value of k is \(\frac{1}{3}\). \[ \boxed{\text{(C) }\dfrac{1}{3}} \]