SOLUTION: A car insurance company has determined that 9% of all drivers were involved
in a car accident last year. Among the 14 drivers living on one particular street, 3
were involved in
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-> SOLUTION: A car insurance company has determined that 9% of all drivers were involved
in a car accident last year. Among the 14 drivers living on one particular street, 3
were involved in
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Question 1188941: A car insurance company has determined that 9% of all drivers were involved
in a car accident last year. Among the 14 drivers living on one particular street, 3
were involved in a car accident last year. If 14 drivers are randomly selected, what
is the probability of getting 3 or more who were involved in a car accident last year? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! p = .09 = probability that a randomly selected driver was involved in an accident last year.
q = 1 - .09 = .91 that a randomly selected driver was not involved in an accident last year.
n = 14 = number of drivers randomly selected.
x = 0 to 14 = number of those drivers that were involved in an accident last year.
p(x) = probability that x number of those 14 drivers that were randomly selected were involved in an accident last year.
p(x) = p^x * q^(n-x) * c(n,x)
c(n,x) = n! / (x! * (n-x)!)
you can find 1 minus (p(0) + p(1) + p(2)) or you can find p(3 to `14).
all of p(x) from x = 0 to 14 is shown in the attached excel spreadsheet.
your solution is that the probability that 3 or more drivers were involved in an accident last year is equal to 0.125510973.
