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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?
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Solution
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
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