sol.hw4.stat3011 - Solutions for Homework 4 Yicheng Kang...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This document was uploaded on 04/28/2011.

Page1 / 4

sol.hw4.stat3011 - Solutions for Homework 4 Yicheng Kang...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online