Answer:
0.9155
Step-by-step explanation:
As, the random variable X the number of accident follows the Poisson distribution so,
P(X=x)=(μ^x)(e^-μ)/x!
μ=average accident=4.1.
The probability that more than one accident occurs per year= P(X>1)=?
P(X>1)=1-P(X≤1)
P(X>1)=1-[P(X=0)+P(X=1)]
P(X=0)=(4.1^0)(e^-4.1)/0!
P(X=0)=0.016573/1
P(X=0)=0.016573
P(X=1)=(4.1^1)(e^-4.1)/1!
P(X=1)=(4.1)0.016573/1
P(X=1)=0.0679481
P(X>1)=1-[0.016573+0.0679481]
P(X>1)=1-0.084521
P(X>1)=0.915479
P(X>1)=0.9155
The probability that more than one accident occurs per year is 91.55%
