Chapter 13.3, Problem 67AYU

To determine

To find: The probability of a household having:

a. 1 or 2 motor vehicles.

Answer to Problem 67AYU

a. $0.71$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

a. $P\left(1or2Vehicles\right)=P\left(1\right)+P\left(2\right)$

$=0.34+0.37=\text{ 0}.71$

To determine

To find: The probability of a household having:

b. 1 or more motor vehicles.

Answer to Problem 67AYU

b. $0.91$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

b. $P\left(1ormorevehicle\right)=P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4ormore\right)$

$=0.34+0.37+0.14+0.06$

$=\text{ 0}.91$

To determine

To find: The probability of a household having:

c. 3 or fewer motor vehicles.

Answer to Problem 67AYU

c. $0.94$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

c. $P\left(3orfewervehicle\right)=P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)$

$=0.09+0.34+0.37+0.14$

$=0.94$

To determine

To find: The probability of a household having:

d. 3 or more motor vehicles.

Answer to Problem 67AYU

d. $0.2$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

d. $P\left(3ormorevehicle\right)=P\left(3\right)+P\left(4ormore\right)$

$=0.14+0.06$

$=0.2$

To determine

To find: The probability of a household having:

e. Fewer than 2 motor vehicles.

Answer to Problem 67AYU

e. $0.43$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

e. $P\left(fewerthan2motorvehicles\right)=P\left(0\right)+P\left(1\right)$

$=0.09+0.34$

$=0.43$

To determine

To find: The probability of a household having:

f. Fewer than 1 motor vehicles.

Answer to Problem 67AYU

f. $0.09$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

f. $P\left(fewerthan1motorvehicles\right)=P\left(0\right)$

$=0.09$

To determine

To find: The probability of a household having:

g. 1, 2, or 3 motor vehicles.

Answer to Problem 67AYU

g. $0.85$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

g. $P\left(1,2or3vehicle\right)=P\left(1\right)+P\left(2\right)+P\left(3\right)$

$=0.34+0.37+0.14$

$=0.85$

To determine

To find: The probability of a household having:

h. 2 or more motor vehicles.

Answer to Problem 67AYU

h. $0.57$

Explanation of Solution

Given:

In a survey about the number of motor vehicles per household, the following probability table was constructed:

Formula Used:

$P\left(AorB\right)=P\left(A\right)+P\left(B\right)$

Calculation:

h. $P\left(2ormorevehicles\right)=P\left(2\right)+P\left(3\right)+P\left(4ormore\right)$

$=0.37+0.14+0.06$

$=0.57$