Question:
A random variable X takes values \(0,1,2,3\), ........ with probabilities. \(\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{2}\right)^x, \mathrm{k}\) is a constant, then \(P(X=1)=\)
A random variable X takes values \(0,1,2,3\), ........ with probabilities. \(\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1)\left(\frac{1}{2}\right)^x, \mathrm{k}\) is a constant, then \(P(X=1)=\)
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Use the condition that the total probability is 1 to find k.
Updated On: Nov 13, 2026
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{8}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The probabilities must add up to 1: \(\sum_{x=0}^{\infty}k(x+1)\left(\dfrac12\right)^x=1\).
Step 2: Sum the Series
\(\sum(x+1)t^x=\dfrac{1}{(1-t)^2}\). With \(t=\dfrac12\) this equals 4.
Step 3: Find k
\(4k=1\Rightarrow k=\dfrac14\).
Step 4: Calculation
\(P(X=1)=\dfrac14\times2\times\dfrac12=\dfrac14\)
Final Answer:
P(X=1) equals 1/4. \[ \boxed{\text{(C) }\dfrac{1}{4}} \]
The probabilities must add up to 1: \(\sum_{x=0}^{\infty}k(x+1)\left(\dfrac12\right)^x=1\).
Step 2: Sum the Series
\(\sum(x+1)t^x=\dfrac{1}{(1-t)^2}\). With \(t=\dfrac12\) this equals 4.
Step 3: Find k
\(4k=1\Rightarrow k=\dfrac14\).
Step 4: Calculation
\(P(X=1)=\dfrac14\times2\times\dfrac12=\dfrac14\)
Final Answer:
P(X=1) equals 1/4. \[ \boxed{\text{(C) }\dfrac{1}{4}} \]
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