TS EAMCET · Maths · Probability
If X is a random variable with probability distribution \(\mathrm{P}(\mathrm{X}=k)=\frac{(2 k+3) \mathrm{c}}{3^k}\), \(k=0,1,2, \ldots\), to \(\infty\), then \(\mathrm{P}(\mathrm{X}=3)=\)
- A \(\frac{1}{24}\)
- B \(\frac{1}{18}\)
- C \(\frac{1}{6}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{18}\)
Step-by-step Solution
Detailed explanation
\(\sum_{k=0}^{\infty} \mathrm{P}(\mathrm{X}=k) = 1 \) \(\mathrm{c} \sum_{k=0}^{\infty} \frac{2 k+3}{3^k} = 1 \) \(\mathrm{c} \left( 2 \sum_{k=0}^{\infty} k \left(\frac{1}{3}\right)^k + 3 \sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k \right) = 1 \)…
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