Chapter 5, Problem 5CRE

a.

To determine

To show:that the probabilities is a legitimate probability distribution.

Explanation of Solution

Given:

Census bureo allows each person to choose from a long list of races.probabilities is a legitimate probability distribution

If it choose U.S. at random the census bureo given these following probabilities:

	Hispanic	Not Hispanic
Asian	0.001	0.044
Black	0.006	0.124
white	0.139	0.674
Others	0.003	0.009

Calculation:

All probabilities have to be between 0 and 1 (including 0 and 1 also).

Thus, the condition is satisfied.

The sum of all probabilities has to be equal to 1:

  $0.001+0.006+0.003+0.044+0.124+0.674+0.009=1$ .

Since both condition are satisfied, the assign of probabilities is a legitimate probability distribution.

b.

To determine

To find:the probabilitythat a randomly chosen American is Hispanic.

Answer to Problem 5CRE

  $P\left(\text{Hispanic}\right)=14.9\%$

Explanation of Solution

Given:

Census bureo allows each person to choose from a long list of races.probabilities is a legitimate probability distribution.

If it choose U.S. at random the census bureo given these following probabilities:

	Hispanic	Not Hispanic
Asian	0.001	0.044
Black	0.006	0.124
white	0.139	0.674
Others	0.003	0.009

Calculation:

The probability of being Hispanic is the sum of all probabilities of being Hispanic for each race:

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

c.

To determine

To find: the probability that a randomly chosen American is not a member of this group.

Answer to Problem 5CRE

  $P\left(\text{Hispanic}\right)=14.9\%$

Explanation of Solution

Given:

  $P\left(\text{Not hispanic and white}\right)=0.674$

Calculation:

Complement rule:

  $P\left(\text{not }A\right)=1-P\left(A\right)$

Then determine the probability of randomly selecting an American that does not belong to the group “Not Hispanic and White”.

  $P\left(\text{not }\left(\text{Not hispanic and white}\right)\right)=1-P\left(\text{Not hispanic and white}\right)=1-0.674=0.326=32.6\%$

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

d.

To determine

To prove: the given equation. Also, to find $P\left(\text{White or Hispanic}\right)$ .

Answer to Problem 5CRE

  $P\left(\text{White or Hispanic}\right)=82.3\%$

Explanation of Solution

Given:

The given equation is “ $P\left(\text{White or Hispanic}\right)\ne P\left(\text{White}\right)+P\left(\text{Hispanic}\right)$

Calculation:

As the events “White and Hispanic are not mutually exclusive, that there are people who can be both white and Hispanic”.

Addition rule:

  $P\left(A\text{or}B\right)=P\left(A\right)+P\left(A\right)-P\left(A\text{ and }B\right)$

The probability of being Hispanic is the sum of all probabilities of being white for and not Hispanic:

  $P\left(\text{Hispanic}\right)=0.001+0.006+0.139+0.003=0.149$ .

The probability of being Hispanic is the sum of all probabilities of being white for Hispanic and not Hispanic:

  $P\left(\text{White}\right)=0.139+0764=0.813$

From table,

  $P\left(\text{White and Hispanic}\right)=0.139$

Thus, it obtained:

  $\begin{array}{l}P\left(\text{White or Hispanic}\right)=P\left(\text{White}\right)+P\left(\text{Hispanic}\right)-P\left(\text{White and Hispanic}\right)-P\left(\text{White and Hispanic}\right)\\ =0.813+0.149-0.139=0.823=82.3\%\end{array}$