91Ó°ÊÓ

Data for African and Asian ancestry in United States:

$82\%$of the population is White

$14\%$ of the population is Black

$4\%$of the population is Asian

Lactose intolerant people:

$15\%$ are Whites

$70\%$are Blacks

$90\%$ are Asians

Probability for White,

$P(\mathrm{W})=0.82$

Probability for Black,

$P\left(B\right)=0.14$

Probability for Asian,

$P(\mathrm{A})=0.04$

Probability for White lactose intolerant,

$P(\mathrm{L}âˆ\pounds \mathrm{W})=0.15$

Probability for Black lactose intolerant,

$P(\mathrm{L}âˆ\pounds \mathrm{B})=0.70$

Probability for Asian lactose intolerant,

$P(\mathrm{L}âˆ\pounds \mathrm{A})=0.90$

Apply general multiplication rule:

Probability for lactose intolerant and White,

$P(\mathrm{L}\text{and}\mathrm{W})=P(\mathrm{W})×P(\mathrm{L}âˆ\pounds \mathrm{W})=0.82×0.15=0.123$

Probability for lactose intolerant and Black,

$P(\mathrm{L}\text{and}\mathrm{B})=P(\mathrm{B})×P(\mathrm{L}âˆ\pounds \mathrm{B})=0.14×0.70=0.098$

Probability for lactose intolerant and Asian,

$P(\mathrm{L}\text{and}\mathrm{A})=P(\mathrm{A})×P(\mathrm{L}âˆ\pounds \mathrm{A})=0.04×0.90=0.036$

Since the group of people considering themselves to belong to more than one race are not included.

Apply general addition rule for mutually exclusive events:

$P(\mathrm{L})=P(\mathrm{L}\text{and}\mathrm{W})+P(\mathrm{L}\text{and}\mathrm{B})+P(\mathrm{L}\text{and}\mathrm{A})$

$=0.123+0.098+0.036$

$=0.257$

Using conditional probability definition:

$P(\mathrm{A}âˆ\pounds \mathrm{L})=\frac{P(\mathrm{L}\text{and}\mathrm{A})}{P(\mathrm{L})}=\frac{0.036}{0.257}=\frac{36}{257}≈0.1401$

Thus,

The conditional probability for randomly selected person is lactose intolerant Asian is approx.$0.1401.$