Suppose for simplicity that the number of children in a familyis 1, 2, or 3, wit

Question

Suppose for simplicity that the number of children in a familyis 1, 2, or 3, with probability 1/3 each. In this setting,answer the following two questions and compare their results.
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(Q1) Little Bobby has no brothers. What is theprobability that he is an only child? . (Q2) Little Bobby has no sisters. What is theprobability that he is an only child? Suppose for simplicity that the number of children in a familyis 1, 2, or 3, with probability 1/3 each. In this setting,answer the following two questions and compare their results.
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(Q1) Little Bobby has no brothers. What is theprobability that he is an only child? . (Q2) Little Bobby has no sisters. What is theprobability that he is an only child?

Explanation / Answer

Outcomes in sample space are For 1child:          g,b                      each probability 1/2 For two children:  gb,bb,gg              each with probability 1/3 For 3 children:      bgg,bbg,bbb,ggg   each with probability 1/4 1) Bobby no brothers: b,bg,bgg P(only child)=1/3 P(bg)=P(bg and 2 children)=P(bg/2)P(2)=(1/3)(1/3)=1/9 P(bgg)=P(bgg and 2 children)=(1/4)(1/3)=1/12 So P(bobby has no brothers)=1/3+1/9+1/12=19/36 and P(only child/no brothers)=P(no brothers/only child)P(onlychild)/P(no brothers) =(1)(1/3)/(19/36)=12/19 2)P(only child /no sisters)=P(no sisters/only child)P(onlychild)/P(no sisters) P(no sisters)=P(no sisters/only child)P(only child)+P(bb/2children)P(2 children)+P(bbb/3 children)P(3 children) =1(1/3)+(1/3)(1/3)+(1/4)(1/3)=19/36 and P(only child/no sisters)=(1)(1/3)/(19/36)=12/19 The probabilites are the same as would be expected.

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