Question

According to company records, 5% of all automobiles brought to Geoff Garage last year for a state-mandated annual inspection did not pass. Of the next 10 automobiles entering the inspection station:

(a) (10 points) what is the probability that more than 3 will NOT pass inspection? (Show the work by using a relevant mathematical formula, DO NOT use Excel formula)

(b) (5 points) what is the probability exactly seven will pass inspection. (Show the work by using relevant mathematical formula, DO NOT use Excel formula)

Answer 1

Solution:
Given in the question
P(Annual inspection did not pass) = 0.05
P(Annual inspection pass) = 1 - 0.05 = 0.95
No. of automobiles entering the inspection station = 10
Here we will use binomial probability distribution because all events are independent of each other and there is only two outcomes of every event, binomial distribution probability can be calculated as
P(X=n|N,p) = NCp*(p^n)*((1-p)^(N-n))
Solution(a)
Here we need to calculate the probability that more than 3 will not pass inspection
P(X>3) =? which can be calculated as
P(X>3) = 1-P(X<=3) = 1 - P(X=0) - P(X=1) -P(X=2) - P(X=3) = 1 - 10C0*(0.05)^0*(0.95)^10 -10C1*(0.05)^1*(0.95)^9 -10C2*(0.05)^2*(0.95)^8 -10C3*(0.05)^3*(0.95)^7 = 1 - 0.5987 - 0.3151 - 0.0746 - 0.0105 = 0.0010
So there is 0.1% probability that more than 3 will not pass inspection.
Solution(b)
Here we need to calculate the probability exactly seven will pass inspection
P(pass inspection) = 0.95
So P(X=7) can be calculated as
P(X=7) = 10C7*(0.95)^7*(0.05)^3 = 120*0.6983*0.000125 = 0.0105
So there is 1.05% probability that exactly seven will pass inspection.