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GrumpyCorp drug tests all of the recent college graduates it hires each year. The drug test currently used correctly determines drug users 96% of the time(a Positive test) and correctly determines non-users 90% of the time(a Negative test). A recent study concluded that 36% of college students use drugs. A potential employee has been tested and the result was Negative for drug use.
a) Construct ALL necessary probabilities using proper notation(Example: P(D) for a "drug user"). (Hint: there should be 6 total)
b) Find the Probability of a Negative test, by showing use of the above Probabilities first, and then followed by the proper calculation.
c) Use Bayes' Theorem to find the probability that a person who tests Positive actually was not a Drug User. Set up using Conditional Probability Notation and then substitute in numeric values.

Respuesta :

Solution :

Drug : Drug user

T : Test positive

a). [tex]P(D) =0.36[/tex]

    [tex]$P\left(\frac{T}{D} \right) = 0.96$[/tex]

    [tex]$P\left(\frac{T^c}{D^c} \right) = 0.90$[/tex]

b). [tex]$P(T^c)= P\left(\frac{T^c}{D^c}\right) \times P(D^c)+ P\left(\frac{T^c}{D}\right) \times P(D)$[/tex]

               [tex]$=0.9 \times (1-0.36) + (1-0.96) \times 0.36$[/tex]

               = 0.5904

c). [tex]$P\left(\frac{D^c}{T}\right) = \frac{P\left(\frac{T}{D^c}\right). P(D^c)}{P\left(\frac{T}{D^c}\right). P(D^c) + P\left(\frac{T}{D}\right). P(D)}$[/tex]

                  [tex]$=\frac{(1-0.90) \times (1-0.36)}{(1-0.90) \times (1-0.36)+(0.96 \times 0.36)}$[/tex]

                 [tex]$=0.15625$[/tex]

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