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.

Answer 1

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%.