Answer:
0.009804
Step-by-step explanation:
We are given;
probability of testing positive given that you have the disease is 0.99
Also, probability of not testing positive and not having the disease is 0.99
We are also told that it is a rare disease and so strikes only 1 in 1000 people = 0.0001
Let's denote positive test by T+, negative test by T¯, having the disease by D+, not having the disease by D¯.
So, we can now denote all the values in probability we have written earlier.
Thus:
P(T+ | D+) = 0.99
P(T¯ | D¯) = 0.99
P(D+) = 0.0001
Thus, P(D¯) = 1 - P(D+) = 1 - 0.0001 = 0.9999
Now, let's find probability of testing positive;
P(T+) = (P(T+ | D+) × P(D+)) + (P(T+ | D¯) × P(D¯))
Now, (P(T+ | D¯) is not given but by inspection, we can infer from the values given that it is 0.01
Thus;
P(T+) = (0.99 × 0.0001) + (0.01 × 0.9999)
P(T+) = 0.010098
Chances that one has the disease would be gotten from Baye's theorem;
P(D+ | T+) = (P(T+ | D+) × P(D+))/P(T+) = (0.99 × 0.0001)/0.010098 = 0.009804
