Question

The probability that a student is in the band, or a freshman, is .25. The probability that a student is a freshman is .31. The probability that a student is in the band is .08. What is the probability that the student is in the band and a freshman?
Hint:

P(A) + P(B) – P(A + B)
Group of answer choices

14%

39%

48%

64%

Answer 1

To find the probability that a student is in the band and a freshman, we can use the formula for the union of two events:

\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

Where:
- \( P(A \cup B) \) is the probability that a student is in the band or a freshman.
- \( P(A) \) is the probability that a student is in the band.
- \( P(B) \) is the probability that a student is a freshman.
- \( P(A \cap B) \) is the probability that a student is both in the band and a freshman.

We are given:
- \( P(A \cup B) = 0.25 \)
- \( P(A) = 0.08 \)
- \( P(B) = 0.31 \)

We need to find \( P(A \cap B) \). Rearranging the formula gives us:

\[
P(A \cap B) = P(A) + P(B) - P(A \cup B)
\]

Substituting the values we have:

\[
P(A \cap B) = 0.08 + 0.31 - 0.25
\]

Calculating this step-by-step:

1. \( 0.08 + 0.31 = 0.39 \)
2. \( 0.39 - 0.25 = 0.14 \)

Therefore, \( P(A \cap B) = 0.14 \).

So, the probability that a student is in the band and a freshman is **14%**.

Thus, the answer is:

**14%**