4-4 Multiplication Rule: Basics Pre-Employment Drug Screening. In Exercises 13-1
ID: 3350950 • Letter: 4
Question
4-4 Multiplication Rule: Basics Pre-Employment Drug Screening. In Exercises 13-16, use the test results summarized in Table 4-1, reproduced here. Consider an event to be "unlikely if its probability is 0.05 or less Table 4-1 Pre-Employment Drug Screening Results Positive Test Result (Drug Use Is Indicated)(Drug Use Is Not Indicated) Negative Test Result Subject Uses Drugs True Positive) 90 (False Positive) False Negative) 860 True Negative) Subject Is Not a Drug User 13. Pre-Employment Drug Screening If 2 of the 1000 test subjects are randomly selected, find the probability that they both had false positive results. Is it unlikely to randomly select 2 subjects and get 2 results that are both false positive results? a. Assume that the 2 selections are made with replacement. b. Assume that the 2 selections are made without replacement. 14. Pre-Employment Drug Screening If 3 of the 1000 test subjects are randomly selected find the probability that they all had false negative results. Is it unlikely to randomly select 3 subjects and get 3 results that are all false negative results? a. Assume that the 3 selections are made with replacement. b. Assume that the 3 selections are made without replacement. 15. Pre-Employment Drug Screening If 3 of the 1000 test subjects are randomly selected, find the probability that they all had correct test results (either true positive or true negative). Is such an event unlikely? a. Assume that the 3 selections are made with replacement b. Assume that the 3 selections are made without replacement.Explanation / Answer
(13)(a) probability of first false positive=90/1000
probability of second false positive=90/1000
required probability=(90/1000)*(90/1000)=0.0081
13(b)probability of first false positive=90/1000
probability of second false positive=89/999
required probability=(90/1000)*(89/999)=0.008018
(14a)probability of first false negative=6/1000
probability of second false negative=6/1000
probability of third false negative=6/1000
required probability=(6/1000)*(6/1000)*(6/1000)=0.000000216
(14b)
probability of first false negative=6/1000
probability of second false negative=5/999
probability of third false negative=4/998
required probability=(6/1000)*(5/999)*(4/998)=0.00000012
(15a)probability of first True results=904/1000
probability of second True results=904/1000
probability of third True results=904/1000
required probability=(904/1000)*(904/1000)*(904/1000)=0.7388
(15b) probability of first True results=904/1000
probability of second True results=903/999
probability of third True results=902/998
required probability=(904/1000)*(903/999)*(902/998)=0.7385
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