2.48
		t =	1	week
		Time Period:	Peak
		Arrival Rate:	37	accidents/year	accidents/year
		Arrival Rate (u):	0.71	accidents/week	accidents/week
	P(No vehicles while backing out) = P(H >= 9)
		P(n)=e-ut(ut)n/n!
P(at least 1) = P(n > 0) = 1 - P(n <=0) = 1 - P(0) = 1 - [e-(0.71acc/week)(1 week)][(0.71 acc/week)(1 week)]0/0!
		P(at least 1) = 1 - 0.492 = 0.508
	What is the probability of at least 2 accidents next week?
P(at least 2) = P(n > 1) = 1 - P(n <=1) = 1 - {P(0) + P(1)} = 1 - {0.492 + [e-(0.71acc/week)(1 week)][(0.71 acc/week)(1 week)]1/1!}
		P(at least 2) = 1 - 0.492 - 0.349 = 0.159
	What is the probability of exactly 4 accidents next week?
P(4) = P(n = 4) = [e-(0.71acc/week)(1 week)][(0.71 acc/week)(1 week)]4/4!
P(4) =[(0.492)(0.254)]4/(1*2*3*4)
		P(4) = 0.005