Bayes' Rule Question

Question

I am reading about Bayes' rule, I can solve all the exercise but this one. Suppose you had a checkup, and there is a bad news; you tested positive for "the giggles" and that the test is 99% accurate( i.e., the probability of testing positive when you do have "the giggles" is 0.99, as is the probability of testing negative when you don't.) The good news is that this rare condition, striking only 1 in 10000 people of your age. Why is good news that the condition is rare? what are the chances that you actually have "the giggles"?

Answer 1

Consider what happens with testing a million people (having the right proportion of sick people): 100 of them have "the giggles" and if we test them we find 99 positive tests, as the test is 99% reliable.

Of the remaining (1000000 - 100) = 999900 non-"giggles" people and 1% of them test positive (as 99% of them test negative), which means 9999 positive tests.

Of all positive tests (9999 + 99 = 10098) only 99 actually have it.

So the probability is $\frac{99}{10098}$, which is less than $0.01$.

Formally: if $G$ means having the "giggles", and $T^+$ is a positive test, $T^-$ is a negative test.

Then $$P(G|T^+) = \frac{P(G \cap T^+)}{P(T^+)} = \frac{P(T^+|G)P(G)}{P(T^+|G)P(G) + P(T^+|\lnot G)P(\lnot G)}$$

Plugging in:

$$ \frac{0.99 \cdot 0.0001}{0.99 \cdot 0.0001 + 0.01 \cdot 0.9999}$$

Which are the exact same numbers as in the numerical example (multiply everything by a million...). This might make the idea a bit more accessible.