One percent of population cannot drive even if they try very very hard, but everyone applies for the driving license. The driving test fails those who cannot drive with a chance of 97%, but because the test has to be strict, it fails those who drive well with a probability of 3%. How likely it is that the person who failed a driving test is actually an able driver?
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4$\begingroup$ Please edit your title to say more about your question since in this form it is meaningless $\endgroup$– TimCommented Jun 29, 2016 at 12:46
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2 Answers
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The topic you are searching for is diagnostic tests. Key concepts are the sensitivity (the proportion of true cases detected) specificity (the proportion of true non-cases detected as such) and positive and negative predictive value. The concepts also appear in other fields under different names. In order to calculate them with a confidence interval you will need the frequencies rather than the percentages.
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$\begingroup$ The work has one exact answer and does not need confidence intervalls. $\endgroup$– BernhardCommented Jun 29, 2016 at 13:12
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For this Kind of question, contingency tables are very helpful. The Information given can be written as the following table:
can drive | cannot drive | sum
pass test : | |
fail in test: 3% of 99% of total | 97% of 1% of total |
sum : 99% of total | 1% of total | 100% of total
From there on you can easily compute the missing cells. First step: compute which percentage of total fails the test. Then you know how many of the total pass the test. Then you go on filling in the upper row and using the upper left cell you can compute the answer.
"This community's policy is to "provide helpful hints" for self-study questions."
Hope, this is a helpful hint. Bernhard
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$\begingroup$ Can I solve this using Bayes' formula? $\endgroup$ Commented Jun 29, 2016 at 20:52
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$\begingroup$ If that is the context in which you were given the task, then that is probably the way to go. Even better, do both ways and see, how they are the same and are not the same.Assign "can drive" the letter A and "passes test" the letter B. Now assign one of the given numbers to P(not B|A) and all other given numbers to the adequate element of Bayes rule. It should burn down to some simple algebra, then. Compare your results to the contingency table. $\endgroup$– BernhardCommented Jun 29, 2016 at 21:46
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$\begingroup$ Is the answer 75% using Bayes' formula? $\endgroup$ Commented Jun 30, 2016 at 21:18