Question 1203189: A recent study of robberies for a certain geographic area showed an average of 1 robbery per 20,000 people. In a city of 80,000 people, find the probability of the following...
a) 0 robberies
b) 1 robbery
c) 2 robberies
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! assuming this is a binomial distribution, i get the following:
n = 80,000
p = 1/20,000 = .00005
q = (1 - 1/20,000) = .99995
x = 0, 1, or 2
formula is p(x) = p^x * q^(n-x) * c(n,x)
we get:
p(0) = .00005^0 * .99995^(80,000 - 0) * c(80,000,0) = 0.018313807.
p(1) = .00005^1 * .99995^(80,000 - 1) * C(80,000,1) = 0.073258892.
P(2) = .00005^2 * .99995^(80,000 - 2) * C(80,000,2) = 0.146523279.
the combination formula is c(n,x) = n! / (x! * (n-x)!)
for example, when n = 80,000 and x = 2, this formula becomes:
c(80,000,2) = 80,000 / (2! * 79,998!) which becomes:
c(80,000,2) = (80,000 * 79,999 * 79,998!) / (2! * 79,998!) which becomes:
c(80,000,2) = (80,000 * 79,999) / 2 = 3,199,960,000.
P(2) = .00005^2 * .99995^(80,000 - 2) * C(80,000,2) = 0.146523279 becomes:
P(2) = .00005^2 * .99995^(80,000 - 2) * 3,199,960,000 = 0.146523279.
you can verify with your calculator that this is correct.
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