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HWKsolutions3lastsemester

# 5andpx005 furtherwehavechosenthesettingofybdietobe

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Unformatted text preview: on of Y == B == Die, we see that the P(Die=3) = is appx 0.24, where the theoretical should have been 0.1667. Empirically: P(A⋂B) = P(B | A) P(A) = (0.24) * (0.5) = 0.12 = P(B) * P(A) Hence P(A⋂B) = P(B) * P(A) as required, specifically, P(B|A) = P(B) 4 http://www.stat.ucla.edu/%7Edinov/courses_students.dir/12/Winter/STAT13.1.dir/HWs.dir/HW03.html The statement P(B|A) = P(B) is true since knowing the outcome of A “The fair coin where p = 0.5” tells us nothing about B “The die roll” since we chose a 1‐6 flat setting – i.e., all the outcomes {1,...,6} are still equally likely (discrete uniformly distributed) Theoretically: P(A⋂B) = P(B|A) P(A) = (0.16667) * (0.5) = 0.0833 = P(B) * P(A) P(B=3) = P(B=3|A=1)P(A=1) + P(B=3|A=0)P(A=0) = 0.0833...
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