3. Suppose that a couple has four children. a) List the sample space of all poss

Question

3. Suppose that a couple has four children. a) List the sample space of all possible outcomes for the sexes of these four children. (For example, three of the possible outcomes are BiB2BsB4, BiG2B:G4, and BiB2GsG4, where Bi means that the h child is a boy and Gi means that the th child is a girl.) Now assume that all sixteen outcomes in the sample space are equally likely. b) Report the probability that all four children are of the same sex. c Which is more likely- that the couple will have 2 children of each sex or a 3-1 breakdown? Justify your answer.

Explanation / Answer

3)

a)sample space of 16 outcome ={ B1B2B3B4 ;B1B2B3G4 ;B1B2G3B4 ;B1G2B3B4 ;G1B2B3B4 ;B1B2G3G4 ;B1G2B3G4 ;G1B2B3G4 ;B1G2G3B4 ;G1B2G3B4 ;G1G2B3B4 ;B1G2G3G4 ;G1B2G3G4 ;G1G2G3B4 ;G1G2B3G4 ;G1G2G3G4 }

b) there are two outcomes for same sex ={ B1B2B3B4 ; G1G2G3G4 }

hence probability =2/16 =1/8

c)

number of outcome of 2 children of same sex ={B1B2G3G4 ;B1G2B3G4 ;G1B2B3G4 ;B1G2G3B4 ;G1B2G3B4 ;G1G2B3B4 } =6

hence probaility of 2 children of same sex = 6/16 =3/8

number of outcome of 3:1 breakdown =

{B1B2B3G4 ;B1B2G3B4 ;B1G2B3B4 ;G1B2B3B4 ;B1G2G3G4 ;G1B2G3G4 ;G1G2G3B4 ;G1G2B3G4 } =8

hence probaility of  outcome of 3:1 breakdown = 6/16 =3/8

from above we can see that probability of couple having 2 children of each sex or a 3:1 breakdown are equal

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