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In a certain Algebra $2$ class of $23$ students, $15$ of them play basketball and $16$ of them play baseball. There are $3$ students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

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Find probabilities using the addition rule

Full solution

Q. In a certain Algebra

2

class of

23

students,

15

of them play basketball and

16

of them play baseball. There are

3

students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

Calculate total students: Calculate the total number of students who play at least one sport. Since $3$ students play neither sport, the number of students playing at least one sport is $23 - 3 = 20$ .
Use inclusion-exclusion principle: Use the principle of inclusion-exclusion to find the number of students who play both sports. Let $A$ be the set of students who play basketball and $B$ be the set of students who play baseball. According to the principle, $|A \cup B| = |A| + |B| - |A \cap B|$ . We know $|A \cup B| = 20$ (from step $1$ ), $|A| = 15$ , and $|B| = 16$ . Plugging in these values, we get $20 = 15 + 16 - |A \cap B|$ . Solving for $|A \cap B|$ , we find $|A \cap B| = 15 + 16 - 20 = 11$ .
Calculate probability: Calculate the probability that a randomly chosen student plays both sports. The probability is the number of students who play both sports divided by the total number of students. Probability = $\frac{|A \cap B|}{\text{Total students}} = \frac{11}{23}$ .

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