Suppose the prior probability of the germ carrier is 10%. When in tests, germ carriers have the probability of $95\%$ to give positive results and $5\%$ to give negative; non-germ carriers have the probability of $1\%$ to give positive results and $99\%$ to give negative. Now, there is a person who passes the test three times, giving $2$ positive and $1$ negative results. The question is to calculate the probability of this person being germ carrier. The tests are independent.
My answer is:
Notate the prior probability of:
germ carrier $ P(A)=10 \% $
non-germ carrier $ P(\bar A)=90 \% $
and the event of
positive $b=1$
negative $b=0$
We list the conditional probability:
$P(b=1|A) = 95\%$
$P(b=0|A) = 5\%$
$P(b=1|\bar{A}) = 1\%$
$P(b=0|\bar{A}) = 99\%$
So, we calculate the prior probability of test results: $$P(b=1)=P(b=1|A)P(A) + P(b=1|\bar{A})P(\bar{A})=10.4\%$$ $$P(b=0)=P(b=0|A)P(A) + P(b=0|\bar{A})P(\bar{A})=89.6\%$$ Finally, we calculate the posterior by knowing the tests are independent
$$ P(A|b=1,1,0)=\frac{P(b=1,1,0|A)P(A)}{P(b=1,1,0)} = \frac{P(b=1|A)P(b=1|A)P(b=0|A)P(A)}{P(b=1)P(b=1)P(b=0)} \approx 46.56\%$$
So, did I do it right?
Let $X$ be the 3 tests.
$P(A|X)=\frac{P(X|A)P(A)}{P(X|\bar{A})P(\bar{A})+P(X|A)P(A)}$
$=\frac{0.1\times C_3^2\times 0.95^2\times 0.05}{0.1\times C_3^2\times 0.95^2\times 0.05+0.9\times C_3^1\times0.01^2\times0.99 }$
$=0.9806$
Am I correct? I need your confirmation.