.
People who drink more than 5 drinks in a week are likely to develop cirrhosis of liver 95% of time,
whereas people who drink fewer than 5 drinks develop cirrhosis of the liver 20% of the time.
About 5% of the population has more than 5 drinks in a week.
If you learn that a person has cirrhosis, what are the odds that they drink more than 5 drinks in a week?
~~~~~~~~~~~~~~~~~~~~~~~~
First of all, I want to notice that the problem's formulation is not fully perfect.
It determines probability for more than 5 drinks and fewer than 5 drinks,
but leaves the gap "at 5 drinks".
So, for correctness, it should be reformulated.
I modify it this way: "for more than 5 drinks" and "for less or equal 5 drink"
to fulfill this gap.
Below is my solution for the MODIFIED formulation.
In this problem, we have these conditional probabilities given (in shorten form)
(C stands for cirrhosis; ds stands for drinks)
(a) P(C | has > 5 ds ) = 0.95;
(b) P(C | (has <= 5 ds ) = 0.2;
(c) P(has > 5 ds) = 0.05 ( which implies P(has <= 5 ds) = 0.95 ).
They want you find P(has > 5 drinks | C).
From (a) we have P(C and has > 5 ds) = 0.95*0.05; (1)
From (b) we have P(C and has <= 5 ds) = 0.2*0.95. (think WHY it is so !)
It implies P(C) = P(C and has > 5 ds) + P(C and has <= 5 ds) = 0.95*0.05 + 0.2*0.95 = 0.2375. (2)
Notice that in (2), I use the modified formulation, which ELIMINATES the gap.
Now the final calculation is
P(C and has > 5 ds)
P(has > 5 drinks | C) = ----------------------- = (from (1) and (2) ) = = 0.2 (rounded). ANSWER
P(C)
Solved.
For the solution, and in order for to understand it, you should have good feeling of conditional probability.
-------------------
It is an advanced problem for those students, who just have an experience
solving simple and typical assignments on conditional probability.
It is intended to separate those students who is able to work at advanced level.