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hw5_418_10

# hw5_418_10 - 58 a S = H,T T,H are the only possible space...

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Unformatted text preview: 58. a) S = { ( H,T ) , ( T,H ) } are the only possible space, since you need to restart if get the same case. So, p ( H ) = p ( TH ) = (1- p ) p , and p ( T ) = p ( HT ) = p (1- p ) . Hence they are the same. When considering the conditional case, p ( H ) = p ( TH | S ) = p ( TH ) p ( S ) = (1 − p ) p 2(1 − p ) p = 1 / 2 . Similarly, p ( T ) = p ( HT | S ) = p ( HT ) p ( S ) = (1 − p ) p 2(1 − p ) p = 1 / 2. b) This simpler procedure cannot guarantee the same probability. For example, you ﬂip TT, then you stop on the third ﬂip if you get H, i.e. TTH, then p ( H ) = (1- p ) 2 p , but if you get T on the third ﬂip, you need to continue. 60. Let B denotes 1 blue gene, R denotes 1 brown gene. Since Smith’s sister has blue eyes, then both of Smith’s parents must be BR. All possible com- bination of offspring for BR × BR is BB (with probability 1/4), BR (with probability 1/2), and RR (with probability 1/4). a) Let brown denotes that Smith has brown eyes, Blue denotes that Smith has a blue gene. Then p ( blue | brown ) = p ( blue ∩ brown ) p ( brown ) = p ( BR ) p ( BR ∪ RR ) = 2 / 4 3 / 4 = 2 / 3 . b) If child has blue eye, then both Smith and his wife need to offer B genes to this child. Since Smith’s wife has blue eyes, her genes must be BB. But for Smith, we only know that he has gene either BR, or RR. Now, for the case of BB (wife) and RR (Smith), they can only offer BR gene to the offspring; for the case of BB (wife) and RB, the genes that they offer to the offspring...
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hw5_418_10 - 58 a S = H,T T,H are the only possible space...

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