Suppose a certain drug test is 91% sensitive, that is, the test will correctly i

Question

Suppose a certain drug test is 91% sensitive, that is, the test will correctly identify a drug user as testing of the time, and 93% positive 91% specific, that is, the test will correctly identify a non-user as testing negative 93% of the time. Suppose a corporation drug. What is the probability that, given a positive drug test, an employee is actually a drug user? Let D stand for be a drug user and N indicate being a non-user. Letbe the event of a positive drug test and e be the event of a negative drug test. decides to test its employees for drug use, and that only 0.4% of the employees actually use the (b) P(N) = (d) P(BIN) = (f) P(DIB) Please answer all parts of the question. Check The submission was invalid, and has been disregarded without penalty Finish attempt

Explanation / Answer

Note that +' = - and D' = N

P(+|D) = 0.91 P(-|N) = 0.93 => P(+'|D') = 0.93

P(D) = 0.004

(b) P(N) = P(D') = 1 - 0.004 = 0.996

(d) P(D and +) = P(+|D) P(D) = 0.91 * 0.004 = 0.00364

P(D' and +') = P(+'|D') P(D') = 0.93 * 0.996 = 0.92628

=> P(D U +) = 1 - P(D' and +') = 1 - 0.92628 = 0.07372.

P(+) = P(D U +) + P(D and +) - P(D)

= 0.07372 + 0.00364 - 0.004

= 0.07336.

P(+ and D') = P(+) - P(+ and D)

= 0.07336 - 0.00364

= 0.06972.

P(+|N) = P(+ and D') / P(D') = 0.06972 / 0.996 = 0.07.

(e) P(+) = 0.07336.

(f) P(D|+) = P(D and +) / P(+)

=  0.00364 / 0.07336

= 0.0496.

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