Asked by Steve
                Twenty tires are tested to see if they last as long as the manufacturer claims they do. Three tires fail the test. Two tires are selected at random without replacement for an inspection.
Find the probability that both tires fail the test.
Which is greater, the probability that the first tire fails and the second one passes? or vice-versa?
            
        Find the probability that both tires fail the test.
Which is greater, the probability that the first tire fails and the second one passes? or vice-versa?
Answers
                    Answered by
            Damon
            
    p(f) = 3/20
p(first) = 3/20 = .15
p(second) = 2/19 = .105
p first and second = .15 * .105 = .0158
first fail, second pass
= 3/20 * 18/19
first pass second fail = 17/20 * 3/19
    
p(first) = 3/20 = .15
p(second) = 2/19 = .105
p first and second = .15 * .105 = .0158
first fail, second pass
= 3/20 * 18/19
first pass second fail = 17/20 * 3/19
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