P0066_Soln - PF ( 29 PMF ( 29 PAF M ( 29 = .5 (29 .6 (29...

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P0066_Soln.doc We want to know the probability that the offspring have asthma. This total probability is the sum of the probability associated with the four possible situations of the parents’ smoking status: PA ( 29 = PF M A ( 29 + P F M A ( 29 + PF M A ( 29 + P F M A ( 29 The multiplication rule can be used to re-express each of the above situations in terms of conditional probabilities. For example: PF M A ( 29 = PF ( 29 PMF ( 29 PAF M ( 29 The values of those probabilities in the new formulation were given in the problem statement. For example:
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Unformatted text preview: PF ( 29 PMF ( 29 PAF M ( 29 = .5 (29 .6 (29 .15 ( 29 You may find it helpful to depict the information provided in the problem in a graphical manner such as: Father? Mother? Mother? Thus, PA ( 29 = .5 (29 .8 (29 .4 (29 + .5 (29 .2 (29 .13 (29 + .5 (29 .4 (29 .05 (29 + .5 (29 .6 (29 .15 (29 = .084 Refer to Rosner problems at end of chapter 3 on pulmonary disease. Answer: C 0.04 0.13 0.05 0.15 0.5 0.5 0.8 0.2 0.4 0.6 Legend: Yes == No ==...
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This note was uploaded on 02/01/2011 for the course BME 335 taught by Professor Dunn during the Spring '10 term at University of Texas at Austin.

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