91Ó°ÊÓ

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:

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