Question #348112

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


1
Expert's answer
2022-06-07T09:02:13-0400

a.


P(X3)=1P(X=0)P(X=1)P(X\ge 3)=1-P(X=0)-P(X=1)

P(X=2)=1e3(3)00!e3(3)11!-P(X=2)=1-\dfrac{e^{-3}(3)^0}{0!}-\dfrac{e^{-3}(3)^1}{1!}

e3(3)22!=0.576810-\dfrac{e^{-3}(3)^2}{2!}=0.576810

b.


P(X3X1)=P(X3X1)P(X1)P(X\ge3|X\ge1)=\dfrac{P(X\ge3\cap X\ge 1)}{P(X\ge 1)}

=P(X3)P(X1)=1e3(3)00!e3(3)11!e3(3)22!1e3(3)00!e3(3)11!=\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!}}

=0.5766100.800852=0.720246=\dfrac{0.576610}{0.800852}=0.720246


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