Question 1102408: The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 3.5 per week. Find the probability of the following events.
A. No accidents occur in one week.
B. 4 or more accidents occur in a week.
C. One accident occurs today.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! P(0)=e-(3.5)*3.5^0/0!; and 0!=1
This is =0.0302
For 4 or more do 1,2,3
P(1)=e^(-3.5)*3.5/1=0.1057.
P(2)=e^(-3.5)*3.5^2/2!=0.1850
P(3)=e^(-3.5)*3.5^3/3!=0.2158
P(0,1,2,3)=0.5367
1-0.5367 is P(4 or more)=0.4633
Today is one day out of 7, so the Poisson parameter is 1/7 of 3.5 or 0.5
P(1)=e-(0.5)*0.5^1/1!=0.3033
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