Question

Given that P(X=2)=9P(X=4)+90P(X=46) for a Poission distribution variate X. Find (i)P(X<2) (ii)P(X>4) (iii)P(X>=1).

Answer 1

Answer on Question #69232 – Math – Statistics and Probability

Question

Given that $P(X = 2) = 9P(X = 4) + 90P(X = 6)$ for a Poisson distribution variate $X$. Find

i) $P(X < 2)$

ii) $P(X > 4)$

iii) $P(X \geq 1)$

Solution

The probability of observing $x$ events in a given interval is given by

$$P(X = x) = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, 3, \dots; \quad \lambda > 0$$

Given $P(X = 2) = 9P(X = 4) + 90P(X = 6)$

$$\begin{array}{l} e^{-\lambda} \frac{\lambda^2}{2!} = 9e^{-\lambda} \frac{\lambda^4}{4!} + 90e^{-\lambda} \frac{\lambda^6}{6!} \\ \frac{90}{720} \lambda^4 + \frac{9}{24} \lambda^2 - \frac{1}{2} = 0 \\ \lambda^4 + 3\lambda^2 - 4 = 0 \\ (\lambda^2 - 1)(\lambda^2 + 4) = 0 \\ \lambda = 1 \text{ or } \lambda = -1 \text{ or } \lambda^2 = -4; \quad \lambda > 0 \\ \end{array}$$

Hence $\lambda = 1$.

i) $P(X < 2) = P(X = 0) + P(X = 1) = e^{-\lambda} \frac{\lambda^0}{0!} + e^{-\lambda} \frac{\lambda^1}{1!} = e^{-1} + e^{-1} = 2e^{-1} \approx 0.735759$

ii) $P(X > 4) = 1 - \left(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)\right) = 1 - \left(e^{-\lambda} \frac{\lambda^0}{0!} + e^{-\lambda} \frac{\lambda^1}{1!} + e^{-\lambda} \frac{\lambda^2}{2!} + e^{-\lambda} \frac{\lambda^3}{3!} + e^{-\lambda} \frac{\lambda^4}{4!}\right) = 1 - \left(e^{-1} + e^{-1} + e^{-1} \frac{1}{2} + e^{-1} \frac{1}{6} + e^{-1} \frac{1}{24}\right) = 1 - \frac{65}{24} e^{-1} \approx 0.003660$

iii) $P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-\lambda} \frac{\lambda^0}{0!} = 1 - e^{-1} \approx 0.632121$

Answer: i) $2e^{-1}$; ii) $1 - \frac{65}{24} e^{-1}$; iii) $1 - e^{-1}$.

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