Answer
a) 0.035
b) 0.14
Step-by-step explanation:
Let J be the event that Joe has the disease.
Let The be the event that Joe's test is positive.
Pr(J) = 1/2%
= 0.5/100 = 0.005
Pr(J') = 99.5%
= 0.995
Pr(T|J) = 98%
= 0.98 since 2% if the time if a person having the disease is omitted ("false negative ")
Pr(T|J') = 3% = 0.03 since there are 3 false positives
a( Pr(T) = us the probability that Joe tests positive
Pr(T) = Pr(T|J)* P(J) + Pr(T|J')*Pr(J')
= (0.98*0.005) + (0.03*0.995)
= 0.00049 + 0.02985
= 0.03475
= 0.035
b) Pr( J|T) = Pr(JnT) / Pr(T)
= (Pr(T|J)*Pr(J)) / Pr(T)
= (0.005*0.98) / 0.035
= 0.0049/0.035
= 0.14