Unformatted text preview: 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 /2n . The reason is that all the 2n sequences are equally likely, and the k number of sequences with exactly k heads is n , i.e., among the n ﬂips, we choose k ﬂips to be k 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 (AB ) = 1 − P (AB ). A: ¯ P (A ∩ B ) P (B ) − P (A ∩ B ) ¯ P (AB ) = = = 1 − P (AB ). P (B ) P (B ) (2) P (A ∩ B C ) = P (AB ∩ C )P (B C ). A: P (A ∩ B C ) = P (A ∩ B ∩ C ) . P (C ) P (AB ∩ C )P (B C ) = P (A ∩ B ∩ C ) P (A ∩ B ∩ C ) P (B ∩ C ) = . P (B ∩ C ) P (C ) P (C ) Problem 3: Independence. If P (AB ) = P (A), prove (1) P (A ∩ B ) = P (A)P (B ). A: P (AB ) = 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: Suppose an urn has r red balls and b black balls. We randomly pick a ball. Then we put 3 balls of the same color back to the urn. Then we randomly pick a ball again. What is the probability that the second pick is red? A: P (2nd = red) = P (1st = red)P (2nd = red1st = red) + P (1st = blue)P (2nd = red1st = blue) r r+2 b r = + r+br+b+2 r+br+b+2 r = . r+b 1 ...
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 Spring '09
 Wu
 Christopher Nolan, Pick, ball, AB, STAT 100A HWII

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