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Unformatted text preview: Solutions for Homework 4 Yicheng Kang October 12, 2010 5.19 a)The sample space for the possible genders of four children is FFFF,FFFM,FFMF,FFMM,FMFF,FMFM,FMMF,FMMM,MMMM,MMMF,MMFM,MMFF,MFMM, MFMF,MFFM,and MFFF. b)There are 16 possible outcomes, only one of which is all girls. The probability of having all girls, therefore, is 1 / 16 = 0 . 0625. Thus, it is fairly unusual to have a family with four children who are all girls. c)For the calculation in (b) to be valid, we have to assume that it is equally likely to have a boy or a girl, and that the gender of each child is independent of the genders of the others. That is, we have to assume that the gender of your previous child has no influence on the gender of your next child. 5.20 Outcomes are not equally likely. This logic assumes that it is equally likely to have each of these outcomes. In fact, it is more likely to have 1,2 or 3 girls than it is to have 0 or 4 girls. See the sample in 5.19, part a. 5.26 (b) P ( F ) = 0 . 3 + 0 . 1 = 0 . 4 P ( W ) = 0 . 3 + 0 . 05 = 0 . 35 (c)The event “F and W” means that they bought from both catalogs. P(F and W)=0.3....
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- Spring '11
- Probability, partner, additional problem