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?

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

Step-by-step 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

f(x) = y

f(-3) = -5

step-by-step explanation: