Answer :
We start by noting that there are a total of [tex]$60$[/tex] tables, of which [tex]$38$[/tex] are round and [tex]$13$[/tex] are by the window. However, since [tex]$6$[/tex] tables are both round and by the window, we must be careful not to count those twice.
Using the principle of inclusion-exclusion, the number of tables that are either round or by the window is calculated by
[tex]$$
|\,\text{Round} \cup \text{Window}| = |\,\text{Round}| + |\text{Window}| - |\,\text{Round} \cap \text{Window}|.
$$[/tex]
Substitute the given values:
[tex]$$
|\,\text{Round} \cup \text{Window}| = 38 + 13 - 6 = 45.
$$[/tex]
Next, the probability that a randomly assigned table is either round or by the window is the number of favorable outcomes divided by the total number of tables:
[tex]$$
\text{Probability} = \frac{45}{60}.
$$[/tex]
Thus, the correct answer is
[tex]$$
\frac{45}{60}.
$$[/tex]>
So, the answer is [tex]$\boxed{\frac{45}{60}}$[/tex], which corresponds to option B.
Using the principle of inclusion-exclusion, the number of tables that are either round or by the window is calculated by
[tex]$$
|\,\text{Round} \cup \text{Window}| = |\,\text{Round}| + |\text{Window}| - |\,\text{Round} \cap \text{Window}|.
$$[/tex]
Substitute the given values:
[tex]$$
|\,\text{Round} \cup \text{Window}| = 38 + 13 - 6 = 45.
$$[/tex]
Next, the probability that a randomly assigned table is either round or by the window is the number of favorable outcomes divided by the total number of tables:
[tex]$$
\text{Probability} = \frac{45}{60}.
$$[/tex]
Thus, the correct answer is
[tex]$$
\frac{45}{60}.
$$[/tex]>
So, the answer is [tex]$\boxed{\frac{45}{60}}$[/tex], which corresponds to option B.
