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Exercise 9.5 · Q6

Q.A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested. If 0.2% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
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The classic "false positive" Bayes' calculation shows that even with a 99%99\%-effective test, a positive result only means a ≈28.4%\approx28.4\% chance of actually having the disease, because the disease is rare.

Bayes' theorem with D=D= "has the disease", D′=D'= "does not have the disease", +=+= "test result is positive":

P(D∣+)=P(D) P(+∣D)P(D) P(+∣D)+P(D′) P(+∣D′)P(D\mid{+})=\frac{P(D)\,P(+\mid D)}{P(D)\,P(+\mid D)+P(D')\,P(+\mid D')}

  1. Given (prior) probabilities.

P(D)=0.2%=0.002,P(D′)=1−0.002=0.998P(D)=0.2\%=0.002,\qquad P(D')=1-0.002=0.998

  1. Given test-accuracy probabilities.

P(+∣D)=99%=0.99(true positive rate)P(+\mid D)=99\%=0.99\quad(\text{true positive rate})

P(+∣D′)=0.5%=0.005(false positive rate)P(+\mid D')=0.5\%=0.005\quad(\text{false positive rate})

  1. Compute the numerator (true positives among the whole population).

P(D)×P(+∣D)=0.002×0.99=0.00198P(D)\times P(+\mid D)=0.002\times0.99=0.00198

  1. Compute the false-positive contribution.

P(D′)×P(+∣D′)=0.998×0.005=0.00499P(D')\times P(+\mid D')=0.998\times0.005=0.00499

  1. Total probability of testing positive (denominator).

P(+)=0.00198+0.00499=0.00697P(+)=0.00198+0.00499=0.00697

  1. Apply Bayes' theorem. P(D∣+)=0.001980.00697P(D\mid{+})=\frac{0.00198}{0.00697} …

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