Question

Only 0.1% of the individuals in a certain population have a particular disease (an incidence rate of 0.001). Of those who have the disease, 95% test positive when a certain diagnostic test is applied. Of those who do not have the disease, 90% test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test: a) Calculate P(positive test) b) Calculate P(has disease | positive test)

Answer 1

We're given

P(has disease) = 0.001

P(positive test | has disease) = 0.95

P(negative test | does not have disease) = 0.90

(a) By the law of total probability,

P(positive test) = P(positive test AND has disease) + P(positive test AND does not have disease)

Recall that the probability of an event A conditioned on another event B is

P(A | B) = P(A AND B) / P(B)

so that

P(positive test) = P(positive test | has disease) * P(has disease) + P(positive test | does not have disease) * P(does not have disease)

P(positive test) = 0.95 * 0.001 + (1 - 0.90) * (1 - 0.001)

P(positive test) ≈ 0.10

(b) From the definition of conditional probability, we hvae

P(A | B) = P(A AND B) / P(B) = P(B | A) P(A) / P(B)

so that

P(has disease | positive test) = P(positive test | has disease) * P(has disease) / P(positive test)

P(has disease | positive test) = 0.95 * 0.001 / 0.10

P(has disease | positive test) ≈ 0.0094

Answer 2

The probability will be "0.017699".

Given:

Incident rate,

• 0.1% or 0.001
• $P(D) = 0.001$
• $P(D')= 1-0.001$        

                   $= 0.999$

• $P(\frac{P}{D} ) = 0.95$

• $P(\frac{N}{D} ) = 0.9$

Now,

→ $P(\frac{N}{D'} ) = 1-P(\frac{N'}{D'} )$

             $= 1-0.95$

             $= 0.05$

then,

→ $P(\frac{D}{N} ) = \frac{Pr(D)\times P(\frac{N}{D} )}{P(N)}$

or,

            $= \frac{Pr(D)\times P(\frac{N}{D} )}{P(D) \ P(\frac{N}{D} )+P(D') \ P(\frac{N}{D'} )}$

By substituting the above values, we get

            $= \frac{0.001\times 0.9}{0.001\times 0.9+0.999\times 0.05}$    

            $= \frac{0.0009}{0.0009+0.04995}$

            $= \frac{0.0009}{0.05085}$

            $= 0.017699$

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