## Elementary Statistics (12th Edition)

a) We have 4 different outcomes: Xx, xX, XY, YX. We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$ and that the son has a disease if it has xY or Yx, which here is not a possible outcome out of the 4 possible, hence $probability=\frac{0}{4}=0.$ b)We have 4 different outcomes: Xx, xX, XY, YX. We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$ and that the daughter has a disease if it has xx, which here is not a possible outcome out of the 4 possible, hence $probability=\frac{0}{4}=0.$ c)We have 4 different outcomes: XX, XY, Xx, Yx, from which the male outcomes are XY and Yx, from which xY is the diseaseful one. We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$ and that 1 possibility out of the 2 is "good", hence $probability=\frac{1}{2}=0.5.$ d)We have 4 different outcomes: XX, XY, Xx, Yx, from which the female outcomes are XX and Xx. xx is the diseaseful one. We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$ and that 0 possibility out of the 2 is "good", hence $probability=\frac{0}{2}=0.$