Question

An individual is selected at random from a community in which 1 percent are afflicted with tuberculosis and is X-rayed to detect the presence of the disease. The probability of a positive X-ray result, given that the person selected is tubercular is .90. The probability of a positive X-ray, given that the person selected in not tubercular is .01. What is the probability that a person will be tubercular given that his X-ray result is positive?

a) We have insufficient information to answer this question.

b) .673

c) .99

d) .476

e) .90

Answer 1

P(affected with tuberculosis) = 0.01

P(positive X ray result | affected with tuberculosis) = 0.9

P(positive X ray result | not affected with tuberculosis) = 0.01

P(positive X ray result) = P(positive X ray result | affected with tuberculosis) * P(affected with tuberculosis) + P(positive X ray result | not affected with tuberculosis) * P(not affected with tuberculosis)

                                      = 0.9 * 0.01 + 0.01 * (1 - 0.01)

                                      = 0.0189

P(affected with tuberculosis | positive X ray result) = P(positive X ray result | affected with tuberculosis) * P(affected with tuberculosis) / P(positive X ray result)

                                                                                 = 0.9 * 0.01 / 0.0189

                                                                                 = 0.4762

Option-D) 0.476