# HW2sol1 - STAT516 Solution to Homework 2 1.4.5: a) Let...

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STAT516 Solution to Homework 2 1.4.5: a) Let U1=(urn 1 chosen), U2=(urn 2 chosen), B=(black ball chosen), W=(White ball chosen). b) P(U1)=1/2=P(U2); P(W|U1)=3/5; P(B|U1)=2/5; P(W|U2)=3/7; P(B|U2)=4/7 c) P(B)=P(B|U1)P(U1)+P(B|U2)P(U2)= (2/5)(1/2)+ (4/7)(1/2)=17/35 1.4.6: P (2 nd spade| 1 st black)= P(1 st black and 2 nd spade) / P(1 st black) =[P(1 st spade and 2 nd spade)+P(1 st club and 2 nd spade)] / P(1 st black) =[(13/52)(12/51)+(13/52)(13/51)]/(26/32) =25/102 1.4.7: a) A and B are mutually exclusive, then P(AB)=0. So, P(B)= P(AUB)-P(A)=0.3 b) A and B are independent, then P(AUB)=P(A)+P(B)-P(A)P(B) So, P(B)=[P(AUB)-P(A)] / [1-P(A)]=0.3/0.5=0.6 1/2 1/2 U1 U2 2/5 3/5 4/7 3/7 B W B W

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1.4.8 Assume n cards and all 2n faces are equally likely to show on top. P(white on bottom | black on top)= P(white on bottom and black on top) / P(black on top) = 50 %*(1/2) / [50%*(1/2)+20%] = 5/9 1.5.2 a) P(W2)=P(W1W2)+P(B1W2)= (4/10)(7/13)+(6/10)(4/13)=2/5 b) P(B1|W2)=P(B1W2)/P(W2)=(6/10)(4/13) / (2/5)=6/13 c)
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## This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at University of California, Berkeley.

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HW2sol1 - STAT516 Solution to Homework 2 1.4.5: a) Let...

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