Lactose intolerance Lactose intolerance causes difficulty in digesting dairy products that contain lactose (milk sugar). It is particularly common among people of African and Asian ancestry. In the United States (not including other groups and people who consider themselves to belong to more than one race), 82% of the population is White, 14% is Black, and 4% is Asian. Moreover, 15% of Whites, 70% of Blacks, and 90% of Asians are lactose intolerant. 22 Suppose we select a U.S. person at random and find that the person is lactose intolerant. What’s the probability that she or he is Asian?

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,

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

Probability for Black lactose intolerant,

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

Probability for Asian lactose intolerant,

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

Apply general multiplication rule:

Probability for lactose intolerant and White,

$P(\mathrm{L}\text{and}\mathrm{W})=P(\mathrm{W})\times P(\mathrm{L}\mid \mathrm{W})=0.82\times 0.15=0.123$

Probability for lactose intolerant and Black,

$P(\mathrm{L}\text{and}\mathrm{B})=P(\mathrm{B})\times P(\mathrm{L}\mid \mathrm{B})=0.14\times 0.70=0.098$

Probability for lactose intolerant and Asian,

$P(\mathrm{L}\text{and}\mathrm{A})=P(\mathrm{A})\times P(\mathrm{L}\mid \mathrm{A})=0.04\times 0.90=0.036$

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:

$P(\mathrm{A}\mid \mathrm{L})=\frac{P(\mathrm{L}\text{and}\mathrm{A})}{P(\mathrm{L})}=\frac{0.036}{0.257}=\frac{36}{257}\approx 0.1401$

Thus,

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