Question

A test for syphilis gives a positive result with probability 0.98 when tested on a person whohas the disease. The test gives a negative result with probability 0.9 when tested on a personwho does not have the disease. Assume that 3% of a society has the disease, and compute thefollowing probabilities in the order they are asked. Note that for each individual there are twodisease status(disease, no disease) and two types of test results(positive, negative).

a) Compute the probability that an individual has the disease and has a positive test result. Explain.
b) Compute the probability that an individual does not have the disease and has a positive test result.
c) What does the sum of two probabilities given above correspond to? Explan.
d) If possible, using the three probabilities computed above compute the probability that an individual with a positive test result has the disease. Give reasons to your computation.

Answer 1

a) P₁ = 0,0294

b) P₂ = 0,097

c) P₃ = 0,1264

P₃ is the total probability of the individual having a positive test result

d) P₄ = 0,232

Explanation:

a) Probability that an individual has the disease and has a positive result is P₁ :

P₁ = The probability of having the disease * Probability of + in the test result

P₁ = 0,03 * 0,98

P₁ = 0,0294

b) the probability that an individual does not have the disease and has a positive test result P₂ is:

P₂ = 0,97 * 0,1

P₂ = 0,097

c) P₃ the sum of the two previous probabilities is:

P₃ = 0,0294 + 0,097

P₃ = 0,1264

P₃ is the total probability of the individual having a positive test result

d) Applying the Bayes theorem we can get the probability that an individual with a positive test result has the disease. (P₄ )

Bayes theorem establishes:

P [ A/B ] = P[B/A] * P(A) / P(B)

In this case

P[ with disease/ given (+)tr]

= P [(+)Test/ with disease]* P (with disease)/ P of test +

P₄ = 0,98 * 0,03 / 0,1264

P₄ = 0,232