# ANS_HW2 - 550.310 HW2 Solutions Please do NOT redistribute...

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550.310 HW2 Solutions Please do NOT redistribute!!! a) P(Smoker recover on A) = 500/1100 = 5/11 P(Smoker recover on B) = 300/700 = 3/7 A comes with a higher probability. b) P(Nonsmoker recover on A) = 600/900 = 2/3 P(Nonsmoker recover on B) = 900/1400 = 9/14 A comes with a higher probability. c) P(Recover on A) = ( 500 + 600 )/2100 = 11/20 P(Recover on B) = ( 300 + 900 )/2100 = 12/20 = 3/5 B comes with a higher probability, even though A beats B on both Smoker and Nonsmoker case. (a) 3 33 23 4 (2 1 ) ) 28 PH H ⎛⎞⎛⎞ + ⎜⎟⎜⎟ ⎝⎠⎝⎠ ≥≥ = = = 3 333 123 7 (1 ) ⎛⎞ ⎛⎞ ⎛⎞ ++ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ≥= = 4 1 ) 8 |1 ) 7 ) 8 H H = = = 4 7 (b) 3 3 2 3 1 ) T ⎛⎞ ⎜⎟ ⎝⎠ = = 3 7 ) PT = 3 1 ) 8 | 1 ) 7 ) 8 T T = = = 3 7

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Dice 1\2 1 2 3 4 5 6 1 2 3 4 5 - - 2 3 4 5 - - - 3 4 5 - - - - 4 5 - - - - - 5 - - - - - - 6 - - - - - - 4 (3 6) 36 P Appears S <= 10 (6 ) 36 PS 4 3 ) 36 (3 | 6) 10 ) 36 P S Appears P Appears S < = = < 2 5 P(4-th have hemophilia first 3 have no hemophilia) = 0.5 4 P(First 3 have no hemophilia) = 0.5 + 0.5 4 Thus P=0.5 4 /(0.5+0.5 4 ) ( 80 60) ( 80) 0.2 ( 60) 0.6 ( 80 60) 0.2 1 ( 80 | 60) ( 60) 0.6 3 PL L L L ≥≥ = = ≥= = = = a) 0.176*0.02 = 0.00352 b) 0.176*0.02 + 0.114*0.04 + 0.078*0.94 = 0.0814 c) 0.00352/0.0814 = 0.04324 0.1*0.02/(0.1*0.02 + 0.2*0.01 + 0.7*0.0025) = 0.3478
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## This note was uploaded on 10/21/2008 for the course MATH 302 taught by Professor Goldberg during the Fall '08 term at Johns Hopkins.

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ANS_HW2 - 550.310 HW2 Solutions Please do NOT redistribute...

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