In Exercises 5.16-5.26, express your probability answers as a decimal rounded to three places.

Housing Units. The U.S. Census Bureau publishes data on housing units in American Housing Survey for the United States. The following table provides a frequency distribution for the number of rooms in U.S. housing units. The frequencies are in thousands.

A housing unit is selected at random. Find the probability that the housing unit obtained has

(a) four rooms.

(b) more than four rooms.

(c) one or two rooms.

(d) fewer than one room.

(e) one or more rooms.

The given statement is:

The U.S. Census Bureau publishes data on housing units in American Housing Survey for the United States. The following table provides a frequency distribution for the number of rooms in U.S. housing units. The frequencies are in thousands.

The total instances where the housing unit obtained has more than four rooms is:

$30,440+27,779+17,868+19,257=95,344$

Let's call the occurrence 'E' where the housing unit obtained has more than four rooms.

The probability that the housing unit obtained has more than four rooms is:

$$\begin{gathered} P\left(E\right)=\frac{95,344}{1,32,418} \\ =0.720 \end{gathered}$$

The total instances where the housing unit obtained has one or two rooms is:

$601+1,404=2,005$

Let's call the occurrence 'E' where the housing unit obtained has one or two rooms.

The probability that the housing unit obtained has one or two rooms is:

$$\begin{gathered} P\left(E\right)=\frac{2005}{1,32,418} \\ =0.015 \end{gathered}$$

The total instances where the housing unit obtained has fewer than one room is:

Let's call the occurrence 'E' where the housing unit obtained has fewer than one room.

The probability that the housing unit obtained has fewer than one room is:

$$\begin{gathered} P\left(E\right)=\frac{0}{1,32,418} \\ =0 \end{gathered}$$

The total instances where the housing unit obtained has one or more rooms are:

$601+1,404+11,433+23,636+30,440+27,779+17,868+19,257=1,32,148$

Let's call the occurrence 'E' where the housing unit obtained has one or more rooms.

The probability that the housing unit obtained has one or more rooms is:

$$\begin{gathered} P\left(E\right)=\frac{1,32,418}{1,32,418} \\ =1 \end{gathered}$$