Answer :
First, note that out of 60 tables, there are 38 round tables and 13 tables by the window. However, 6 tables are both round and by the window. To avoid counting these tables twice, we use the inclusion-exclusion principle.
The number of tables that are either round or by the window is given by:
[tex]$$
\text{Union} = (\text{round tables}) + (\text{window tables}) - (\text{round tables by the window})
$$[/tex]
Substitute the given numbers:
[tex]$$
\text{Union} = 38 + 13 - 6 = 45
$$[/tex]
Thus, there are 45 tables that are either round or by the window.
The probability that a customer will be seated at one of these tables is:
[tex]$$
\text{Probability} = \frac{\text{Union}}{\text{Total tables}} = \frac{45}{60} = 0.75
$$[/tex]
Expressed as a fraction, the answer is [tex]$\frac{45}{60}$[/tex], which corresponds to option B.
The number of tables that are either round or by the window is given by:
[tex]$$
\text{Union} = (\text{round tables}) + (\text{window tables}) - (\text{round tables by the window})
$$[/tex]
Substitute the given numbers:
[tex]$$
\text{Union} = 38 + 13 - 6 = 45
$$[/tex]
Thus, there are 45 tables that are either round or by the window.
The probability that a customer will be seated at one of these tables is:
[tex]$$
\text{Probability} = \frac{\text{Union}}{\text{Total tables}} = \frac{45}{60} = 0.75
$$[/tex]
Expressed as a fraction, the answer is [tex]$\frac{45}{60}$[/tex], which corresponds to option B.
