To solve this problem, we can use the principle of Inclusion-Exclusion, which ensures that we don't double-count tables when they fall into both categories (round and window tables).
Let's break it down step by step.
1. Identify the total number of tables:
[tex]\[
\text{Total tables} = 60
\][/tex]
2. Identify the number of round tables:
[tex]\[
\text{Round tables} = 38
\][/tex]
3. Identify the number of window tables:
[tex]\[
\text{Window tables} = 13
\][/tex]
4. Identify the number of tables that are both round and located by the window:
[tex]\[
\text{Round and window tables} = 6
\][/tex]
Using the principle of Inclusion-Exclusion:
[tex]\[
\text{Number of tables that are round or by the window} = (\text{Number of round tables}) + (\text{Number of window tables}) - (\text{Number of round and window tables})
\][/tex]
Substitute the given numbers into the formula:
[tex]\[
\text{Number of tables that are round or by the window}= 38 + 13 - 6 = 45
\][/tex]
Next, we calculate the probability:
[tex]\[
\text{Probability} = \frac{\text{Number of tables that are round or by the window}}{\text{Total number of tables}}
\][/tex]
Plug in the numbers:
[tex]\[
\text{Probability} = \frac{45}{60}
\][/tex]
Simplify the fraction:
[tex]\[
\frac{45}{60} = \frac{3}{4} = 0.75
\][/tex]
Converting to a fraction that matches the multiple-choice options, we get:
[tex]\[
\frac{3}{4} = \frac{45}{60}
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\frac{45}{60}}
\][/tex]
Therefore, the correct option is:
[tex]\[
D. \frac{45}{60}
\][/tex]
