Chapter 4.3, Problem 31E

Blood Types and Rh Factors In addition to being grouped into four types, human blood is grouped by its Rhesus (Rh) factor. Consider the figures below which show the distributions of these groups for Americans.

Choose one American at random. Find the probability that the person

a. Is a universal donor, i.e., has O-negative blood

b. Has type O blood given that the person is Rh+

c. Has A+ or AB− blood

d. Has Rh− given that the person has type B

To determine

To obtain: The probability that the person has O– blood group.

Answer to Problem 31E

The probability that the person has O– blood group is 0.06.

Explanation of Solution

Given info:

The data shows that the distributions of types of blood groupsand Rh factors for Americans.

Calculation:

The total number of Rh factors is shown in the Table (1).

	O	A	B	AB	Total
Rh+	37	34	10	4	85
Rh–	6	6	2	1	15
Total	43	40	12	5	100

Table (1)

Let event A denote that the person has O– blood group.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 6 for ‘Number of outcomes in A’ and 100 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{6}{100}\\ =0.06\end{array}$

Thus, the probability that the person has O– blood group is 0.06.

To determine

To obtain: The probability that person has type O blood given that the person has Rh+ factor.

Answer to Problem 31E

The probability that the person has type O blood given that the person has Rh+ factoris 0.435.

Explanation of Solution

Calculation:

Let event B denote that thethe person has Rh+ factorand event C denote that the person has type O blood.

The probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 85 for ‘Number of outcomes in B’ and 100 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(B\right)=\frac{85}{100}\\ =0.85\end{array}$

The formula for probability of event B and C is,

$P\left(B\text{ and }C\right)=\frac{\text{Number of outcomes in }B\text{ and }C}{\text{Total number of outcomes in the sample space}}$

Substitute 37 for ‘Number of outcomes in B and C’ and 100 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(B\text{ and }C\right)=\frac{37}{100}\\ =0.37\end{array}$

Conditional rule:

The formula for probability of C given B is, $P\left(C|B\right)=\frac{P\left(B\text{ and }C\right)}{P\left(B\right)}$

Substitute 0.82 for $P\left(B\right)$ and 0.37 for $P\left(B\text{ and }C\right)$

$\begin{array}{c}P\left(C|B\right)=\frac{P\left(B\text{ and }C\right)}{P\left(B\right)}\\ =\frac{0.37}{0.85}\\ =0.435\end{array}$

Thus, the probability that the person has type O blood given that the person has Rh+ factor is 0.435.

To determine

To obtain: The probability that person has type A+ blood or the person hasAB–.

Answer to Problem 31E

The probability that person has type A+ blood or the person hasAB– is 0.35.

Explanation of Solution

Calculation:

Let event D denote that the the person hasA+ blood and event E denote that the person has type AB–blood. Also, the event E does not affected by the event D.

The probability of event D is,

$P\left(D\right)=\frac{\text{Number of outcomes in }D}{\text{Total number of outcomes in the sample space}}$

Substitute 34 for ‘Number of outcomes in D’ and100 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(D\right)=\frac{34}{100}\\ =0.34\end{array}$

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of outcomes in }E}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes in C’ and 100 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(E\right)=\frac{1}{100}\\ =0.01\end{array}$

Addition Rule:

The formula for probability of getting event A or event B is,

$P\left(D\text{ or }E\right)=P\left(D\right)+P\left(E\right)-P\left(D\text{ and }E\right)$

Substitute 0.34 for $P\left(D\right)$,0.01for $P\left(D\right)$ and 0 for $P\left(D\text{ and }E\right)$ in $P\left(D\text{ or }E\right)$

$\begin{array}{c}P\left(D\text{ or }E\right)=0.34+0.01-0\\ =0.35\end{array}$

Thus, theprobability that person has type A+ blood or the person hasAB– is 0.35.

To determine

To obtain: The probability that person has type Rh– factor given that the person hasB blood.

Answer to Problem 31E

The probability that person has type Rh– factor given that the person hasB blood is 0.167.

Explanation of Solution

Calculation:

Let event F denote that the the person hasB bloodand event G denote that the person hasRh– factor.

The probability of event F is,

$P\left(F\right)=\frac{\text{Number of outcomes in }F}{\text{Total number of outcomes in the sample space}}$

Substitute 12 for ‘Number of outcomes in F’ and 100 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(F\right)=\frac{12}{100}\\ =0.12\end{array}$

The formula for probability of event F and G is,

$P\left(F\text{ and }G\right)=\frac{\text{Number of outcomes in }F\text{ and }G}{\text{Total number of outcomes in the sample space}}$

Substitute 2 for ‘Number of outcomes in F and G’ and 100 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(F\text{ and }G\right)=\frac{2}{100}\\ =0.02\end{array}$

Conditional rule:

The formula for probability of G given F is, $P\left(G|F\right)=\frac{P\left(G\text{ and }F\right)}{P\left(F\right)}$

Substitute 0.12 for $P\left(F\right)$ and 0.02for $P\left(F\text{ and }G\right)$

$\begin{array}{c}P\left(G|F\right)=\frac{P\left(G\text{ and }F\right)}{P\left(F\right)}\\ =\frac{0.02}{0.12}\\ =0.167\end{array}$

Thus, the probability that the person has type Rh– factor given that the person hasB blood is 0.167.