Question

ATCs are required to undergo periodic random drug testing.
A simple, low-cost urine test is used for initial screening. It has been reported that this particular test has a sensitivity and specificity of 0.96 and 0.93. This means that if there is drug use, the test will detect it 96% of the time. If there is no drug use, the test will be negative 93% of the time.

Based on historical results, the FAA reports that the probability of drug use at a given time is approximately 0.007 (this is called the prevalence of drug use).

1. Draw a probability tree for the situation. (Use this order for the tree: Drug Use(yes/no) -> Test Result(+/-) -> Joint Probabilites)
2. A positive test result puts the air traffic controller’s job in jeopardy. What is the probability of a positive test result?
3. Find the probability an air traffic control truly used drugs, given that the test is positive.

Answer 1

P( drug used ) = 0.007 ; P( drug not used ) = 1 - 0.007 = 0.993

P( Test positive | drug used ) = 0.96 ; P( Test negative | drug used ) = 1 - 0.96 = 0.04

P( Test negative | drug not used ) = 0.93 ;  P( Test positive | drug not used ) = 1 - 0.93 = 0.07

From the given information we can draw the tree diagram as follows

1) A positive test result puts the air traffic controller’s job in jeopardy. What is the probability of a positive test result?

P ( test is positive ) = P( drug used ) * P( Test positive | drug used ) + P( drug not used ) *  P( Test positive | drug not used )

= ( 0.007* 0.96 ) + ( 0.993*0.07 )

= 0.0762

2) Find the probability an air traffic control truly used drugs, given that the test is positive.

P( drug used | test positive )

=

= ( 0.007* 0.96 ) / 0.0762

= 0.0882