Question

The director of student health decides to test every student on campus for tuberculosis. If there are 3000 students out of which only 50 have tuberculosis and the probability of a false positive on the test is 0.008 and a false negative is 0.15, what is the probability that a person who tested positive actually has TB?

Answer 1

Answer: the probability that a person who tested positive actually has TB is 0.64

Explanation:

lets say

T : has tuberculosis

+ : test results +ve

- : test result is -ve

so given that

P(+ITc) = 0.008

P(-ITc) = 0.15

P(T) = 50/3000 = 1/60

P(Tc) = 1 - P(T) = 1 - 1/60 = 59/60

Now required probability

= P(TI+)

P(T∩+) / P(+)

= P(T) × P(+IT) / [[P(T) × P(+IT)] + [(P(Tc) × P(+ITc)]

= P(T) × {1-P(-IT)] / [[P(T) × [1-P(-IT)] + [(P(Tc) × P(+ITc)]

WE SUBSTITUTE

= { 1/60 × (1-0.15) } / [1/60 × (1 - 0.15)] + [(59/60) × 0.008]

= (0.0167 × 0.85) / [(0.0167 × 0.85) + (0.9833 × 0.008)]

= 0.01417 / 0.02206

= 0.6423 ≈ 0.64

∴ the probability that a person who tested positive actually has TB is 0.64