Question

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?

Answer 1

Let event A₁: Disease in serious form,

event A₂: Disease in mild form

event A₃: Subject does not have disease,

event B: Subject tests positive.

P(A₁) = 0.02, P(A₂) = 0.1, P(A₃) = 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