Question

Of all registered automobiles in a certain state. 10% violate the state emissions standard. Twelve automobiles are selected at random to undergo an emission test. How to find the probability that exactly three of them violate the standard?

b)find the probability that fewer than three of them violate the standard
c)find the probability that none of them violate the standard

Answer 1

`"a) "0.08523`
`"b) "0.88913`
`"c) "0.28243`

Explanation:

`"We have a binomial distribution with n=12, p=0.1."`
`"a) "C(12,3) * 0.1^3 * 0.9^9 = 220 * 0.001 * 0.38742 = 0.08523`
`"with "C(n,k) = (n!)/((n-k)! k!)" (combinations)"`
`"b) "0.9^12 + 12 * 0.1 * 0.9^11 + 66 * 0.1^2 * 0.9^10"`
`= 0.9^10 * (0.9^2 + 12*0.1*0.9 + 66*0.1^2)`
`= 0.9^10 * (0.81 + 1.08 + 0.66)`
`= 0.9^10 * 2.55`
`= 0.88913`
`"c) "0.9^12 = 0.28243`