TS EAMCET · Maths · Probability
If the range of a random variable \(X\) is \(\{0,1,2,3,4, \ldots \ldots\}\) with \(P(X=k)=\frac{(k+1) a}{3^k}\) for \(k \geq 0\), then \(a\) is equal to
- A \(\frac{2}{3}\)
- B \(\frac{4}{9}\)
- C \(\frac{8}{27}\)
- D \(\frac{16}{81}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{9}\)
Step-by-step Solution
Detailed explanation
Given that \(P(X=k)=\frac{(k+1) a}{3^k} \text { for } x \in(0,1,2, \ldots \infty)\) As we know that \(P(0)+P(1)+P(2)+\ldots \infty=1\) \(\Rightarrow \quad a+\frac{2 a}{3}+\frac{3 a}{3^2}+\ldots \infty=1\) ...(i)…
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