in a certain algebra 2 class of 30 students, 7 of them play basketball and 19 of them play baseball. there…

in a certain algebra 2 class of 30 students, 7 of them play basketball and 19 of them play baseball. there are 5 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

in a certain algebra 2 class of 30 students, 7 of them play basketball and 19 of them play baseball. there are 5 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

Answer

Answer:

$\frac{7}{10}$

Explanation:

Step1: Use the inclusion - exclusion principle

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$ Here, $n(A)$ is the number of basketball - playing students ($n(A) = 7$), $n(B)$ is the number of baseball - playing students ($n(B)=19$), and $n(A\cap B)$ is the number of students who play both sports ($n(A\cap B) = 5$). So, $n(A\cup B)=7 + 19-5=21$.

Step2: Calculate the probability

The probability $P(A\cup B)$ of choosing a student who plays basketball or baseball is given by the formula $P(A\cup B)=\frac{n(A\cup B)}{n(S)}$, where $n(S)$ is the total number of students in the class ($n(S)=30$). $P(A\cup B)=\frac{21}{30}=\frac{7}{10}$.