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

Answer 1

The responsess to the given question can be defined as follows:

Explanation:

`P(medication\ used) = 0.007 \\\\P( drug \\ot\ used ) = 1 - 0.007 = 0.993\\\\P(test\ positive | medication \ used) = 0.96 \\\\ P(test \ negative | medication \ used) = 1 - 0.96 = 0.04\\\\`

`P(Point | Non-used\ medication )=0.93\\\\ P(Point |No\ medication ) = 1 - 0.93 = 0.07`

As per the given info we draw the tree diagram which is defined in the attachment file.

`P ( test \ is\ positive ) = P( medication \ used ) * P( Test \ positive | medication \ used ) + P( medication not \ used ) * P( Test\ positive | medication \ not \ used )`

`= ( 0.007 * 0.96 ) + ( 0.993* 0.07 )\\\\= 0.0762`

`P( medication \ used | test\ positive )= \frac{\textup{P( medication used ) *P( Test positive given medication used )}}{\textup{P( medication used )}}`

`= (( 0.007* 0.96 ))/(0.0762)\\\\= 0.0882`