Use the table to answer the question.

Question

Soccer Baseball Basketball TOTAL
15 and Under 16 20 8 44
Over 15 17 38 24 79
TOTAL 33 58 32 123
A survey was conducted at a local sports club about the participation of children in various sports. The children were split into two age groups, 15 and under and over 15, and they recorded the number of children who play soccer, baseball, and basketball. What is the probability that a student chosen at random is over 15 or plays basketball? Enter your answer as a fraction; you do not need to reduce the fraction.

(1 point)
The probability that the student is over 15 or plays basketball is
.

Answer 1

To find the probability that a student chosen at random is over 15 or plays basketball, we can use the principle of inclusion-exclusion.

Let's define:

• \( A \) = the event that a student is over 15
• \( B \) = the event that a student plays basketball

First, we need to find \( P(A) \), \( P(B) \), and \( P(A \cap B) \):

1. Total number of students, \( N \): 123

2. Number of students over 15, \( |A| \): 79

3. Number of students playing basketball, \( |B| \): 32

4. Number of students over 15 who play basketball, \( |A \cap B| \): 24

Now, we can calculate the probabilities:

• \( P(A) = \frac{|A|}{N} = \frac{79}{123} \)
• \( P(B) = \frac{|B|}{N} = \frac{32}{123} \)
• \( P(A \cap B) = \frac{|A \cap B|}{N} = \frac{24}{123} \)

Using the principle of inclusion-exclusion: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substituting the values: \[ P(A \cup B) = \frac{79}{123} + \frac{32}{123} - \frac{24}{123} \] Now, combine the fractions: \[ P(A \cup B) = \frac{79 + 32 - 24}{123} = \frac{87}{123} \]

Thus, the probability that a student chosen at random is over 15 or plays basketball is: \[ \frac{87}{123} \]

The final answer is: \(\frac{87}{123}\)