TS EAMCET · Maths · Probability
In a Poisson distribution, if \(\frac{\mathrm{P}(\mathrm{X}=5)}{\mathrm{P}(\mathrm{X}=2)}=\frac{1}{7500}\) and \(\frac{\mathrm{P}(\mathrm{X}=5)}{\mathrm{P}(\mathrm{X}=3)}=\frac{1}{500}\), then the mean of the distribution is
- A \(\frac{1}{15}\)
- B \(\frac{1}{5}\)
- C \(\frac{1}{25}\)
- D \(\frac{1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{5}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{P}(\mathrm{X}=5)}{\mathrm{P}(\mathrm{X}=2)} = \frac{e^{-\lambda}\lambda^5/5!}{e^{-\lambda}\lambda^2/2!} = \frac{\lambda^3}{5 \times 4 \times 3} = \frac{\lambda^3}{60}\) \(\frac{\lambda^3}{60} = \frac{1}{7500}\) \(\lambda^3 = \frac{60}{7500} = \frac{1}{125}\)…
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