Question

A survey conducted at a local high school asked students whether they participate in sports, music programs, or both. The results showed that 60% of the surveyed students participate in sports, 40% participate in music programs, and 20% participate in both sports and music programs.

What is the probability that a student chosen at random from those surveyed participates in either sports or music programs?

Question 5 options:

0.8


0.6


0.4


1

Answer 1

To find the probability that a student participates in either sports or music programs, we can use the principle of inclusion-exclusion.

Let:
- \( P(S) \) be the probability of participating in sports,
- \( P(M) \) be the probability of participating in music programs, and
- \( P(S \cap M) \) be the probability of participating in both.

According to the survey results:
- \( P(S) = 0.60 \) (60% participate in sports),
- \( P(M) = 0.40 \) (40% participate in music programs),
- \( P(S \cap M) = 0.20 \) (20% participate in both).

We want to find \( P(S \cup M) \), the probability of participating in either sports or music programs, which is given by the formula:
\[
P(S \cup M) = P(S) + P(M) - P(S \cap M)
\]

Substituting the values:
\[
P(S \cup M) = 0.60 + 0.40 - 0.20
\]
\[
P(S \cup M) = 1.00 - 0.20
\]
\[
P(S \cup M) = 0.80
\]

Therefore, the probability that a student chosen at random from those surveyed participates in either sports or music programs is **0.8**.

The correct answer is **0.8**.