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In a certain Algebra 2 class of 29 students, 6 of them play basketball and 19 of them play baseball. There are 6 students who play neither sport. What is the probability that a student chosen randomly from the class plays basketball or baseball?
Answer:

In a certain Algebra $2$ class of $29$ students, $6$ of them play basketball and $19$ of them play baseball. There are $6$ students who play neither sport. What is the probability that a student chosen randomly from the class plays basketball or baseball? $\newline$ Answer:

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Find probabilities using the addition rule

Full solution

Q. In a certain Algebra

2

class of

29

students,

6

of them play basketball and

19

of them play baseball. There are

6

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

\newline

Answer:

Define Events: Let's denote the events as follows: $\newline$ A: The student plays basketball. $\newline$ B: The student plays baseball. $\newline$ N: The student plays neither sport. $\newline$ We are given the following information: $\newline$ Total number of students in the class $T$ = $29$ $\newline$ Number of students who play basketball $A$ = $6$ $\newline$ Number of students who play baseball $B$ = $19$ $\newline$ Number of students who play neither sport $N$ = $6$ $\newline$ We need to find the probability that a student chosen randomly from the class plays basketball or baseball. This can be found by subtracting the probability of a student playing neither sport from $1$ , since the probability of playing basketball or baseball is the complement of the probability of playing neither. $\newline$ First, we calculate the probability of a student playing neither sport: $\newline$ $P(N)$ = Number of students who play neither sport / Total number of students $\newline$ $P(N)$ = $29$ $1$ $\newline$ Now, we calculate the probability of a student playing basketball or baseball: $\newline$ $29$ $2$ = $29$ $3$ $\newline$ $29$ $2$ = $29$ $5$ $\newline$ Let's perform the calculation.
Given Information: Performing the calculation from the previous step: $\newline$ $P(A \text{ or } B) = 1 - \frac{6}{29}$ $\newline$ $P(A \text{ or } B) = \frac{29}{29} - \frac{6}{29}$ $\newline$ $P(A \text{ or } B) = \frac{29 - 6}{29}$ $\newline$ $P(A \text{ or } B) = \frac{23}{29}$ $\newline$ This is the probability that a student chosen randomly from the class plays basketball or baseball.

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