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Question
In a town, 10% of the population is HIV positive. A new diagnostic kit arrived in the market. This kit correctly identities HIV positive individual 95% of the time and HIV negative individual 89% of time. A particular patient is tested by this kit and is found to be positive. Then the probability that the patient is actually positive .
- 0.4896
- 0.4896
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Solution
The correct option is A 0.4896
Let A= {Patient is found +ve}
E1={HIV +ve}
E2={HIV −ve}
P(E1)=10 % = 0.1 & P(E2)=0.9
P(A/E1)=P(kit gives correct result for +ve individual) = 0.95
P(A/E2)=P(kit gives wrong result for -ve individual) = 0.11
By Baye's theorem
P(Actually +ve)=P(E1/A)
=P(E1)P(A/E1)P(E1)P(A/E1)+P(E2)P(A/E1)
=0.4896
Let A= {Patient is found +ve}
E1={HIV +ve}
E2={HIV −ve}
P(E1)=10 % = 0.1 & P(E2)=0.9
P(A/E1)=P(kit gives correct result for +ve individual) = 0.95
P(A/E2)=P(kit gives wrong result for -ve individual) = 0.11
By Baye's theorem
P(Actually +ve)=P(E1/A)
=P(E1)P(A/E1)P(E1)P(A/E1)+P(E2)P(A/E1)
=0.4896
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