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### solution1

Course: STAT 309, Fall 2008
School: Wisconsin
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Word Count: 222

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309 Solution STAT/MATH 1 1 1.2.2 (a): P({1,2})=P({1})+P({2})=1/8+1/8=1/4 (b): P({1,2,3})=P({1})+P({2})+P({3})=1/8+1/8+1/8=3/8 (c): P(A)=1/2 if and only if |A| = 4. So the number is 8 = 70 4 2 1.2.3 P({2})=P({1,2})-P({1})=2/3-1/2=1/6 3 1.2.4 P ({2, 3}) = P ({2}) + P ({3}) = 0.5 + 0.3 = 0.8 = 0.7 So the probability is not valid 4 1.2.9 P (S) = P (sS {s}) = sS No! impossible. If we assume P({s})=0 for...

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309 Solution STAT/MATH 1 1 1.2.2 (a): P({1,2})=P({1})+P({2})=1/8+1/8=1/4 (b): P({1,2,3})=P({1})+P({2})+P({3})=1/8+1/8+1/8=3/8 (c): P(A)=1/2 if and only if |A| = 4. So the number is 8 = 70 4 2 1.2.3 P({2})=P({1,2})-P({1})=2/3-1/2=1/6 3 1.2.4 P ({2, 3}) = P ({2}) + P ({3}) = 0.5 + 0.3 = 0.8 = 0.7 So the probability is not valid 4 1.2.9 P (S) = P (sS {s}) = sS No! impossible. If we assume P({s})=0 for every s in S, then P ({s}) = 0 Note, the second equation is because S is nite or countable. However P (S) = 1. This is a contradiction. The assumption P({s})=0 is not true. 5 1.2.10 Yes! For example, Uniform [0,1] distribution. P({s})=0 for every s in [0,1]. 6 By inclusion-exclusion the principle. P (A B) = P (A) + P (B) P (A B). So P (A B) = 2/3 + 1/4 3/4 = 1/6. We have P (A B c ) = P (A) P (A B) = 2/3 1/6 = 1/2 P (Ac B) = P (B) P (A B) = 1/4 1/6 = 1/12 1 7 By the inclusion-exclusion principle. P (A B) = P (A) + P (B) P (A B) = 0.2 + 0.15 P (A B) . Note 0 P (A B) min(P (A), P (B)) = 0.15. So the smallest value of P (A B) = 0.2 when ...

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