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UCSB
PSTAT 5A -
Lecture 3 Additional Questions
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IndependenceLecture 301/10/2013(Lecture 3) 01/10/2013 1 / 3Question:Suppose a family has two children (as usual each birth is independent,and each equally likely to be a boy/girl). We define two events:A = { the family has at least one boy and at least one girl }.B = { the family has at most one girl }.Are these two events independent?(Lecture 3) 01/10/2013 2 / 3Solution:The events are NOT independent. Since we can show thatP(A and B) 6= P(A) · P(B).1P(A) = 1/2 (because Ω = {BB, GB, BG, GG})2P(B) = 3/4 (because Ω = {BB, GB, BG, GG})3So we have: P(A) · P(B) = 1/2 × 3/4 = 3/84We also have: P(A and B) = P(A) = 1/25Therefore, P(A and B) 6= P(A) · P(B)(Lecture 3) 01/10/2013 3 /
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