SOLUTION: Twenty tires are tested to see if they last as long as they claim they do. Three tire fail the test. Two tires are selected at random without replacement for a detained inspection.
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Question 1133728: Twenty tires are tested to see if they last as long as they claim they do. Three tire fail the test. Two tires are selected at random without replacement for a detained inspection.
A. What the probability that both tires failed the test.
B. What is greater the probability that the first tire passed the test and the second one failed or the probability that the first one passed and the second one passed?
C. Suppose 2 tires are selected at random without replacement 60 times, how can we predict the number of times both tires fail the test Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! probability both failed is (3/20)(2/19)=6/380 or 3/190.
first one passed and second failed is 17/20*3/19 or 51/380
both passed is 17/20*16/19=272/380. This is more than 5 times likelier to occur.
The probability is 3/190 and 60 times it is done.
The expected value is 180/190 or about 1 time.
The variance is 180/190*(187/190)=0.9324 with sd's being sqrt of that or 0.97.
