TS EAMCET · Maths · Probability
If a Poisson variate \(X\) satisfies the relation \(P(X=3)=P(X=5)\), then \(P(X=4)=\)
- A \(\frac{50}{3 e^{\sqrt{20}}}\)
- B \(\frac{20000}{3 e^{20}}\)
- C \(\frac{125}{3 e^{10}}\)
- D \(\frac{25}{3 e^{\sqrt{20}}}\)
Answer & Solution
Correct Answer
(A) \(\frac{50}{3 e^{\sqrt{20}}}\)
Step-by-step Solution
Detailed explanation
\( \frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^5}{5!} \) \( \frac{\lambda^3}{6} = \frac{\lambda^5}{120} \) \( \lambda^2 = \frac{120}{6} = 20 \) \( \lambda = \sqrt{20} \) \( P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} = \frac{e^{-\sqrt{20}} (\sqrt{20})^4}{24} \)…
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