TS EAMCET · Maths · Probability
If \(\mathrm{X}\) is a Poisson variate satisfying the condition \(3 P(x=2)=P(x=4)\) then \(P(x-6)=\)
- A \(\frac{162}{5 e^6}\)
- B \(\frac{108}{5 \mathrm{e}^6}\)
- C \(\frac{324}{5 \mathrm{e}^6}\)
- D \(\frac{648}{5 e^6}\)
Answer & Solution
Correct Answer
(C) \(\frac{324}{5 \mathrm{e}^6}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {} 3 P(x=2)=P(x=4) \\ & \because P(x)=\frac{e^{-\lambda} \lambda^x}{x !} \\ & \Rightarrow \frac{3 \cdot e^{-\lambda} \lambda^2}{2 !}=\frac{e^{-\lambda} \lambda^4}{4 !} \\ & \Rightarrow \lambda^2=\frac{4 ! \times 3}{2 !}=36 \\ & \Rightarrow \lambda=6 \\ &…
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