in a certain algebra 2 class of 30 students, 7 of them play basketball and 11 of them play baseball. there…

in a certain algebra 2 class of 30 students, 7 of them play basketball and 11 of them play baseball. there are 3 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 11 of them play baseball. there are 3 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

Answer

Answer:

$\frac{15}{30}=\frac{1}{2}$

Explanation:

Step1: Use the formula for $P(A\cup B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Step2: Calculate $P(A)$, $P(B)$ and $P(A\cap B)$

$P(A)=\frac{7}{30}$, $P(B)=\frac{11}{30}$, $P(A\cap B)=\frac{3}{30}$

Step3: Substitute values into the formula

$P(A\cup B)=\frac{7}{30}+\frac{11}{30}-\frac{3}{30}=\frac{7 + 11- 3}{30}=\frac{15}{30}$