Answer to Question #348112 in Statistics and Probability for yousef

Question

suppose that no of accident occuring on high way is possion with random vairables lamda=3

Answer 1

a.

$$P(X\ge 3)=1-P(X=0)-P(X=1)$$

$$-P(X=2)=1-\dfrac{e^{-3}(3)^0}{0!}-\dfrac{e^{-3}(3)^1}{1!}$$

$$-\dfrac{e^{-3}(3)^2}{2!}=0.576810$$

b.

$$P(X\ge3|X\ge1)=\dfrac{P(X\ge3\cap X\ge 1)}{P(X\ge 1)}$$

$$=\dfrac{P(X\ge3)}{P(X\ge 1)}=\dfrac{1-\dfrac{e^{-3}(3)^0}{0!}-\dfrac{e^{-3}(3)^1}{1!}-\dfrac{e^{-3}(3)^2}{2!}}{1-\dfrac{e^{-3}(3)^0}{0!}-\dfrac{e^{-3}(3)^1}{1!}}$$

$$=\dfrac{0.576610}{0.800852}=0.720246$$