Question

A random variable \(X\) takes the values \(0,1,2,3\), \(\qquad\) with probability

\(\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{5}\right)^x\), where k is a constant.

Then \(\mathrm{P}(\mathrm{X}=0)\) is

• A. \(\frac{16}{25}\)
• B. \(\frac{7}{25}\)
• C. \(\frac{19}{25}\)
• D. \(\frac{18}{25}\)

Hint:

Use the sum of probabilities equal to 1 with the series for 1/(1 - r)^2.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The probabilities must add up to 1, so \(k\sum_{x\ge0}(x+1)\left(\frac15\right)^x=1\).

Step 2: Sum the series
\(\sum_{x\ge0}(x+1)r^x=\frac{1}{(1-r)^2}\), with \(r=\frac15\) this is \(\frac{1}{(4/5)^2}=\frac{25}{16}\).

Step 3: Find k
\(k\cdot\frac{25}{16}=1\Rightarrow k=\frac{16}{25}\).

Step 4: Find P(X = 0)
\(P(X=0)=k(0+1)\cdot1=\frac{16}{25}\).

Final Answer:
\(P(X=0)=\frac{16}{25}\). \[ \boxed{\text{(A) }\dfrac{16}{25}} \]