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100AHW2S

# 100AHW2S - STAT 100A HWII Solution Problem 1 If we ip a...

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STAT 100A HWII Solution Problem 1: If we flip a fair coin n times independently, what is the probability that we observe k heads? k = 0 , 1 , ..., n . Please explain your answer. A: The probability is ( n k ) / 2 n . The reason is that all the 2 n sequences are equally likely, and the number of sequences with exactly k heads is ( n k ) , i.e., among the n flips, we choose k flips to be heads, and the rest of the n - k flips to be tails. The number of such choices is ( n k ) . Problem 2: Prove the following two identities: (1) P ( ¯ A | B ) = 1 - P ( A | B ). A: P ( ¯ A | B ) = P ( ¯ A B ) P ( B ) = P ( B ) - P ( A B ) P ( B ) = 1 - P ( A | B ) . (2) P ( A B | C ) = P ( A | C ) + P ( B | C ) - P ( A B | C ). A: P ( A B | C ) = P (( A B ) C ) P ( C ) = P (( A C ) ( B C )) P ( C ) = P ( A C ) + P ( B C ) - P (( A C ) ( B C )) P ( C ) = P ( A | C ) + P ( B | C ) - P ( A B | C ) Problem 3: Independence. If P ( A | B ) = P ( A ), prove (1) P ( A B ) = P ( A ) P ( B ). A: P ( A | B ) = P ( A B ) /P ( B ) = P ( A ), so P ( A B ) = P ( A ) P ( B ). (2) P ( B | A ) = P ( B ). A: P ( B | A ) = P ( B A ) /P ( A ) = P ( A ) P ( B ) /P ( A ) = P ( B ). Problem 4: Prove the following two identities ( C stands for cause, E stands for effect). (1) Rule of total probability: P ( E ) = P ( E | C ) P ( C ) + P ( E | ¯ C ) P ( ¯ C ). A: P ( E ) = P ( E C ) + P ( E ¯ C ) = P ( C ) P ( E | C ) + P ( ¯ C ) P ( E | ¯ C ). (2) Bayes rule: P ( C | E ) = P ( C ) P ( E | C ) / [ P ( C ) P ( E | C ) + P ( ¯ C ) P ( E | ¯ C ]. A: P ( C | E ) = P ( C E ) /P ( E ) = P ( C ) P ( E | C ) / [ P ( C ) P ( E | C ) + P ( ¯ C ) P ( E | ¯ C ]. Problem 5: Suppose 1% of the population is inflicted with a particular disease. For a medical test,
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