Lesson Upper league problems on conditional probability

Upper league problems on conditional probability

Problem 1

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)


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) 

Problem 2

This problem is to determine the conditional probability.


If a student did not study the subject, the probability for him to answer 8 of 10 
questions correctly is

    P(did not study) =  =  = 0.043945313

(the standard formula for the binomial distribution).



If a student did study the subject, the probability for him to answer 8 of 10 
questions correctly is

    P(did study) =  =  = 0.301989888

(the standard formula for the binomial distribution).



Thus the probability for a random student to answer 8 of 10 questions is

    P = 0.5*P(did not study) + 0.5*(did study) = 0.5*0.043945313 + 0.5*0.301989888 = 0.1729676.


We use the weights 0.5, because Only half of all students study for the exam.


Therefore, the conditional probability that Alice studied for exam given that she answered correctly 8 questions is

      =  = 0.8730  (rounded).    ANSWER