2% of the population have a certain blood disease of a serious form: 10% have it in a mild form; and 88% don't have it at all. A new blood test is developed; the probability of testing positive is `9/10` if the subject has the serious form, `6/10` if the subject has the mild form, and `1/10` if the subject doesn't have the disease. A subject is tested positive. What is the probability that the subject has serious form of the disease?

#### Solution

Let event A_{1}: Disease in serious form,

event A_{2}: Disease in mild form

event A_{3}: Subject does not have disease,

event B: Subject tests positive.

P(A_{1}) = 0.02, P(A_{2}) = 0.1, P(A_{3}) = 0.88

The probability of testing positive is `9/10` if the subject has the serious form, `6/10` if the subject has the mild form, and `1/10` if the subject doesn’t have the disease.

∴ `"P"("B"/"A"_1)` = 0.9, `"P"("B"/"A"_2)` = 0.6, `"p"("B"/"A"_3)` = 0.1

P(B) = `"P"("A"_1) "P"("B"/"A"_1) + "P"("A"_2) "P"("B"/"A"_2) + "P"("A"_3) "P"("B"/"A"_3)`

= 0.02 × 0.9 + 0.1 × 0.6 + 0.88 × 0.1

= 0.166

Required probability = `"P"("A"_1/"B")`

By Baye’s theorem

`"P"("A"_1/"B") = ("P"("A"_1) "P"("B"/"A"_1))/("P"("B"))`

= `(0.9 xx 0.02)/0.166`

= 0.108