To find the probability that a student chosen at random is over 15 or plays basketball, we can use the formula for the probability of the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
where:
- \(A\) is the event that a student is over 15.
- \(B\) is the event that a student plays basketball.
To solve this, we first need the following information:
- Total number of students surveyed.
- Number of students over 15.
- Number of students who play basketball.
- Number of students who are both over 15 and play basketball.
Let's denote:
- Total number of students = \(N\)
- Students over 15 = \(N_{15+}\)
- Students who play basketball = \(N_{B}\)
- Students over 15 who play basketball = \(N_{15+ \cap B}\)
Now we can calculate each component:
- \(P(A) = \frac{N_{15+}}{N}\)
- \(P(B) = \frac{N_{B}}{N}\)
- \(P(A \cap B) = \frac{N_{15+ \cap B}}{N}\)
Plugging these into the formula:
\[ P(A \cup B) = \frac{N_{15+}}{N} + \frac{N_{B}}{N} - \frac{N_{15+ \cap B}}{N} \]
This simplifies to:
\[ P(A \cup B) = \frac{N_{15+} + N_{B} - N_{15+ \cap B}}{N} \]
Since we do not have the specific numbers from the survey, we cannot compute an exact fraction. You would need to substitute the actual values from the survey data into the equation to get the final probability as a fraction.
If you provide the necessary data (the counts of students in each category), I can help you calculate the probability accurately. Otherwise, the final answer for \(P(A \cup B)\) will depend on the specific values collected in your survey.