Answer :
To solve this problem, we need to find the probability that a customer will be seated at either a round table or by the window. We can calculate this using the formula for the probability of the union of two events:
1. Identify the events:
- Let Event A be the event of sitting at a round table.
- Let Event B be the event of sitting at a table by the window.
2. Find the probabilities:
- Total number of tables = 60.
- Number of round tables (Event A) = 38.
- So, the probability of a round table, [tex]\( P(A) \)[/tex], is [tex]\( \frac{38}{60} \)[/tex].
- Number of window tables (Event B) = 13.
- So, the probability of a window table, [tex]\( P(B) \)[/tex], is [tex]\( \frac{13}{60} \)[/tex].
- Number of tables that are both round and by the window (A and B) = 6.
- So, the probability of both round and window, [tex]\( P(A \cap B) \)[/tex], is [tex]\( \frac{6}{60} \)[/tex].
3. Use the probability formula for the union of two events:
[tex]\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\][/tex]
- Substituting the values in, we get:
[tex]\[
P(A \cup B) = \frac{38}{60} + \frac{13}{60} - \frac{6}{60}
\][/tex]
- Add and subtract the fractions:
[tex]\[
P(A \cup B) = \frac{38 + 13 - 6}{60} = \frac{45}{60}
\][/tex]
4. Simplify the fraction:
- The result can be simplified by finding the greatest common divisor of 45 and 60, which is 15:
[tex]\[
\frac{45}{60} = \frac{45 \div 15}{60 \div 15} = \frac{3}{4} = 0.75
\][/tex]
Therefore, the probability that a customer will be seated at a round table or by the window is [tex]\(\frac{45}{60}\)[/tex], which equals 0.75. So, the correct answer is (D) [tex]\(\frac{45}{60}\)[/tex].
1. Identify the events:
- Let Event A be the event of sitting at a round table.
- Let Event B be the event of sitting at a table by the window.
2. Find the probabilities:
- Total number of tables = 60.
- Number of round tables (Event A) = 38.
- So, the probability of a round table, [tex]\( P(A) \)[/tex], is [tex]\( \frac{38}{60} \)[/tex].
- Number of window tables (Event B) = 13.
- So, the probability of a window table, [tex]\( P(B) \)[/tex], is [tex]\( \frac{13}{60} \)[/tex].
- Number of tables that are both round and by the window (A and B) = 6.
- So, the probability of both round and window, [tex]\( P(A \cap B) \)[/tex], is [tex]\( \frac{6}{60} \)[/tex].
3. Use the probability formula for the union of two events:
[tex]\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\][/tex]
- Substituting the values in, we get:
[tex]\[
P(A \cup B) = \frac{38}{60} + \frac{13}{60} - \frac{6}{60}
\][/tex]
- Add and subtract the fractions:
[tex]\[
P(A \cup B) = \frac{38 + 13 - 6}{60} = \frac{45}{60}
\][/tex]
4. Simplify the fraction:
- The result can be simplified by finding the greatest common divisor of 45 and 60, which is 15:
[tex]\[
\frac{45}{60} = \frac{45 \div 15}{60 \div 15} = \frac{3}{4} = 0.75
\][/tex]
Therefore, the probability that a customer will be seated at a round table or by the window is [tex]\(\frac{45}{60}\)[/tex], which equals 0.75. So, the correct answer is (D) [tex]\(\frac{45}{60}\)[/tex].
