Race and ethnicity The Census Bureau allows each person to choose from a long list of races. That is, in the eyes of the Census Bureau, you belong to whatever race you say you belong to. Hispanic (also called Latino) is a separate category. Hispanics may be of any race. If we choose a resident of the United States at random, the Census Bureau gives these probabilities:$25$

(a) Verify that this is a legitimate assignment of probabilities.

(b) What is the probability that a randomly chosen American is Hispanic?

(c) Non-Hispanic whites are the historical majority in the United States. What is the probability that a randomly chosen American is not a member of this group?

(d) Explain why P(white or Hispanic) ≠ P(white) +P(Hispanic). Then find P(white or Hispanic).

• Use a Venn diagram to model a chance process involving two events.
• Use the general addition rule to calculate P(A ∪ B).

Each person can choose from a vast list of races when filling out the census bureau. A genuine probability distribution is probabilities.

If the census bureau chose the United States at random, the following probability would apply:

Addition rule: $P(AorB)=P\left(A\right)+P\left(A\right)-P(AandB)$

All probabilities must be in the range of $0$to $1$(including $0$ and $1$also).

As a result, the criterion is met.

The total number of probability must equal one: $0.001+0.006+0.003+0.044+0.124+0.674+0.009=1$

Because both conditions are met, assigning probabilities is a valid probability distribution.

The total of all Hispanic probabilities for each race is the probability of being Hispanic: $P\left(Hispanic\right)=0.001+0.006+0.139+0.003=0.149=14.9\%$

$P(Nothispanicandwhite)=0.674$

Complement rule: $P(notA)=1-P\left(A\right)$

Complement rule: $P(notA)=1-P\left(A\right)$

Then calculate the likelihood of selecting an American who is not Hispanic or White at random. $$\begin{gathered} P(not(NotHispanicandwhite\left)\right)=1-P(NotHispanicandwhite) \\ =1-0.674 \\ =0.326 \\ =32.6\% \end{gathered}$$

$P\left(Hispanic\right)=0.001+0.006+0.139+0.003=0.149=14.9\%$

The equation is $P(WhiteorHispanic)\ne P\left(White\right)+P\left(Hispanic\right)$

"White and Hispanic are not mutually exclusive, that there are persons who can be both white and Hispanic," according to the events.

Addition rule: $P(AorB)=P\left(A\right)+P\left(A\right)-P(AandB)$

The sum of all probabilities of being white for and not being Hispanic is the chance of being Hispanic: $P\left(Hispanic\right)=0.001+0.006+0.139+0.003=0.149$

For Hispanics and non-Hispanics, the likelihood of being Hispanic equals the sum of the probabilities of being white:$P\left(White\right)=0.139+0764=0.813$

from the table $P(WhiteandHispanic)=0.139$

As a result, the following was obtained:

$=0.813+0.149-0.139=0.823=82.3\%$