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In a certain Algebra 2 class of 29 students, 18 of them play basketball and 6 of them play baseball. There are 7 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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In a certain Algebra $2$ class of $29$ students, $18$ of them play basketball and $6$ of them play baseball. There are $7$ students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball? $\newline$ Answer:

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Probability of independent and dependent events

Full solution

Q. In a certain Algebra

2

class of

29

students,

18

of them play basketball and

6

of them play baseball. There are

7

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

\newline

Answer:

Calculate Total Students Playing Sports: Let's first determine the total number of students who play at least one sport. Since there are $7$ students who play neither sport, the number of students who play at least one sport is the total number of students minus those who play neither. $\newline$ Calculation: $29$ students (total) - $7$ students (play neither) = $22$ students (play at least one sport).
Apply Inclusion-Exclusion Principle: Now, let's use the principle of inclusion-exclusion to find out how many students play both sports. The principle states that the number of students playing either basketball or baseball is equal to the sum of students playing each sport minus the number of students playing both sports. $\newline$ Calculation: Let $B$ be the number of students playing basketball, let $Ba$ be the number of students playing baseball, and let $B \cap Ba$ be the number of students playing both. We have $B = 18$ , $Ba = 6$ , and we need to find $B \cap Ba$ . $\newline$ According to the principle of inclusion-exclusion: $\newline$ $B + Ba - B \cap Ba =$ Number of students playing at least one sport. $\newline$ $18 + 6 - B \cap Ba = 22$ .
Find Students Playing Both Sports: Solving the equation from the previous step for $B\cap Ba$ gives us the number of students playing both basketball and baseball. $\newline$ Calculation: $18 + 6 - B\cap Ba = 22$ $\newline$ $24 - B\cap Ba = 22$ $\newline$ $B\cap Ba = 24 - 22$ $\newline$ $B\cap Ba = 2$ .
Calculate Probability: Finally, we calculate the probability that a student chosen randomly from the class plays both basketball and baseball. The probability is the number of students playing both sports divided by the total number of students. $\newline$ Calculation: Probability = $B\cap Ba / \text{Total number of students}$ $\newline$ Probability = $\frac{2}{29}$ .

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