Question
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Part 1
In a certain Algebra and Trigonometry class, there are 11 male freshmen, 15 female freshmen, 11 male sophomores, and 15 female sophomores. If a person is selected randomly from the group, find the probability that the selected person is a freshman or female.
Part 1
In a certain Algebra and Trigonometry class, there are 11 male freshmen, 15 female freshmen, 11 male sophomores, and 15 female sophomores. If a person is selected randomly from the group, find the probability that the selected person is a freshman or female.
Answers
GPT-4o mini
To find the probability that a randomly selected person is a freshman or female, we can use the formula for the probability of the union of two events:
\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \]
Where:
- \( A \) is the event that the selected person is a freshman.
- \( B \) is the event that the selected person is female.
**Step 1: Determine the total number of students.**
- Male freshmen: 11
- Female freshmen: 15
- Male sophomores: 11
- Female sophomores: 15
Total number of students:
\[
11 + 15 + 11 + 15 = 52
\]
**Step 2: Calculate \( P(A) \)** (the probability of selecting a freshman).
The total number of freshmen (male and female):
\[
11 + 15 = 26
\]
So, the probability \( P(A) \) is:
\[
P(A) = \frac{26}{52} = \frac{1}{2}
\]
**Step 3: Calculate \( P(B) \)** (the probability of selecting a female).
The total number of females (freshmen and sophomores):
\[
15 + 15 = 30
\]
So, the probability \( P(B) \) is:
\[
P(B) = \frac{30}{52} = \frac{15}{26}
\]
**Step 4: Calculate \( P(A \text{ and } B) \)** (the probability of selecting a freshman who is female).
The total number of female freshmen:
\[
15
\]
So, the probability \( P(A \text{ and } B) \) is:
\[
P(A \text{ and } B) = \frac{15}{52}
\]
**Step 5: Use the formula to find \( P(A \text{ or } B) \)**.
Now, we substitute these values into the formula:
\[
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
\]
\[
P(A \text{ or } B) = \frac{1}{2} + \frac{15}{26} - \frac{15}{52}
\]
To do this calculation, we can convert \( \frac{1}{2} \) and \( \frac{15}{26} \) to have a common denominator of 52:
\[
\frac{1}{2} = \frac{26}{52} \quad \text{and} \quad \frac{15}{26} = \frac{30}{52}
\]
Now we can substitute:
\[
P(A \text{ or } B) = \frac{26}{52} + \frac{30}{52} - \frac{15}{52}
\]
Combining the fractions:
\[
P(A \text{ or } B) = \frac{26 + 30 - 15}{52} = \frac{41}{52}
\]
**Conclusion:**
The probability that a randomly selected person is a freshman or female is
\[
\frac{41}{52}
\] or approximately 0.7885, which is about 78.85%.
\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \]
Where:
- \( A \) is the event that the selected person is a freshman.
- \( B \) is the event that the selected person is female.
**Step 1: Determine the total number of students.**
- Male freshmen: 11
- Female freshmen: 15
- Male sophomores: 11
- Female sophomores: 15
Total number of students:
\[
11 + 15 + 11 + 15 = 52
\]
**Step 2: Calculate \( P(A) \)** (the probability of selecting a freshman).
The total number of freshmen (male and female):
\[
11 + 15 = 26
\]
So, the probability \( P(A) \) is:
\[
P(A) = \frac{26}{52} = \frac{1}{2}
\]
**Step 3: Calculate \( P(B) \)** (the probability of selecting a female).
The total number of females (freshmen and sophomores):
\[
15 + 15 = 30
\]
So, the probability \( P(B) \) is:
\[
P(B) = \frac{30}{52} = \frac{15}{26}
\]
**Step 4: Calculate \( P(A \text{ and } B) \)** (the probability of selecting a freshman who is female).
The total number of female freshmen:
\[
15
\]
So, the probability \( P(A \text{ and } B) \) is:
\[
P(A \text{ and } B) = \frac{15}{52}
\]
**Step 5: Use the formula to find \( P(A \text{ or } B) \)**.
Now, we substitute these values into the formula:
\[
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
\]
\[
P(A \text{ or } B) = \frac{1}{2} + \frac{15}{26} - \frac{15}{52}
\]
To do this calculation, we can convert \( \frac{1}{2} \) and \( \frac{15}{26} \) to have a common denominator of 52:
\[
\frac{1}{2} = \frac{26}{52} \quad \text{and} \quad \frac{15}{26} = \frac{30}{52}
\]
Now we can substitute:
\[
P(A \text{ or } B) = \frac{26}{52} + \frac{30}{52} - \frac{15}{52}
\]
Combining the fractions:
\[
P(A \text{ or } B) = \frac{26 + 30 - 15}{52} = \frac{41}{52}
\]
**Conclusion:**
The probability that a randomly selected person is a freshman or female is
\[
\frac{41}{52}
\] or approximately 0.7885, which is about 78.85%.