After your yearly checkup, the doctor has some bad news and some good news. The bad news is that you tested positive for a serious disease, and the test is 99% accurate( i. e. that probability of testing positive given that you have the disease is .99, as is the probability of testing negative given that you don’t have the disease). The good news is that this is a rare disease, striking only one in 10,000 people. What are the changes that you actually have the disease?

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

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2