Question

Given that P(X=2)=8P(X=4)+80P(X=46) for a Poisson distribution variate X. Find

1) P(X<2)

2) P(X>4)

3) P(X>=1)

Answer 1

For Poisson's distribution, $P(X=x)$  is given by 

$$P(X=x)=\dfrac{e^{-\lambda}\cdot\lambda^x}{x!}$$

It is given that $P(X=2)=8P(X=4)+80P(X=6)$

$$\dfrac{e^{-\lambda}\cdot\lambda^2}{2!}=8\cdot\dfrac{e^{-\lambda}\cdot\lambda^4}{4!}+80\cdot\dfrac{e^{-\lambda}\cdot\lambda^6}{6!}$$

$$\dfrac{\lambda^2}{2}=\dfrac{\lambda^4}{3}+\dfrac{\lambda^6}{9}$$

Since $\lambda>0$ we obtain

$$2\lambda^4+6\lambda^2-9=0$$

$$\lambda^2=\dfrac{-3+3\sqrt{3}}{2}, \lambda>0$$

$$\lambda=\sqrt{\dfrac{-3+3\sqrt{3}}{2}}\approx1.0479$$

1)

$$P(X<2)=P(X=0)+P(X=1)$$

$$=\dfrac{e^{-1.0479}\cdot(1.0479)^0}{0!}+\dfrac{e^{-1.0479}\cdot(1.0479)^1}{1!}$$

$$\approx0.71814$$

2)

$$P(X>4)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)=$$

$$=1-\dfrac{e^{-1.0479}\cdot(1.0479)^0}{0!}-\dfrac{e^{-1.0479}\cdot(1.0479)^1}{1!}$$

$$-\dfrac{e^{-1.0479}\cdot(1.0479)^2}{2!}-\dfrac{e^{-1.0479}\cdot(1.0479)^3}{3!}$$

$$\approx0.00445$$

3)

$$P(X\ge1)=1-P(X=0)$$

$$=1-\dfrac{e^{-1.0479}\cdot(1.0479)^0}{0!}\approx0.64933$$