# 100AHW3 - STAT 100A HWIII Solution Problem 1 Prove the...

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STAT 100A HWIII Solution Problem 1: Prove the following two identities ( C stands for cause, E stands for eﬀect), where { C i ,i = 1 ,...,n } partition the whole sample space. (1) Rule of total probability: P ( E ) = n i =1 P ( C i ) P ( E | C i ). A: P ( E ) = P ( n i =1 ( E C i )) = n i =1 P ( E C i ) = n i =1 P ( C i ) P ( E | C i ). (2) Bayes rule: P ( C j | E ) = P ( C j ) P ( E | C j ) / n i =1 P ( C i ) P ( E | C i ). A: P ( C j | E ) = P ( C j E ) /P ( E ) = P ( C j ) P ( E | C j ) / n i =1 P ( C i ) P ( E | C i ). Problem 2: Suppose 1% of the population is inﬂicted with a particular disease. For a medical test, if a person has the disease, then 95% chance the person will be tested positive. If a person does not have the disease, then 90% chance the person will be tested negative. Using precise notation, calculate (1) The probability that a randomly selected person will be tested positive. A: Let
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