Question

The incidence of a deadly disease, among a certain population, is 0.01%. Individuals, randomly selected from this population are submitted to a test whose accuracy is 99% both ways. That is to say, the proportion of positive results among people known to be affected by the disease is 99%. Likewise, testing people that are not suffering from the disease yields 99% negative results. The test gives independent results when repeated. An individual test positive.

a) What is the probability that this person is actually affected?

b) How much more likely are they to have SIDA if they test positive? How likely overall are they if they test positive?

c) The test is repeated twice. What is the probability that the person has SIDA if all three tests are positive? If the first is positive and the next two tests are negative?

Answer 1

P(positive) = P(positive | affected)*P(affected) + P(positive | not affected)*P(not affected)

= 0.99 * (0.0001) + (1 - 0.99) * (1 - 0.0001)

= 0.010098

a.

P(affected | positive) = P(positive | affected)*P(affected) / P(positive)

= 0.99 * (0.0001) / 0.010098

= 0.009804

P(affected | positive) = 0.009804

b.

P(affected | positive) = 0.009804

while P(affected) = 0.01% = 0.0001

P(affected | positive) / P(affected)

= 98.04

therefore the person is 98.04 times more likely to be affected if they test positive

likeliness if they test positive = P(affected | positive) = 0.009804

c.

if all three tests are positive :

P(3 positive) = (P(positive | affected) ^ 3)*P(affected) + (P(positive | not affected) ^ 3)*P(not affected)

= (0.99)^3 * (0.0001) + (1 - 0.99)^3 * (1 - 0.0001)

= 0.0000980298

P(affected | 3 positive) = P(positive | affected)^3 *P(affected) / P(3 positive)

= (0.99)^3 * (0.0001) / 0.0000980298

= 0.9898

If the first is positive and the next two tests are negative :

P(1 positive, 2 negative) = P(positive | affected)*P(negative | affected)^2 *P(affected) + P(positive | not affected)*P(negative | not affected)^2 *P(not affected)

= (0.99)* (1-0.99)^2 * (0.0001) + (1 - 0.99)* (0.99)^2 * (1 - 0.0001)

= 0.0098000298

P(affectedd | 1 positive, 2 negative) = P(positive | affected)*P(negative | affected)^2 *P(affected) / P(1 positive, 2 negative)

= (0.99)* (1-0.99)^2 * (0.0001) / 0.0098000298

= 0.0000010102

(PLEASE UPVOTE)