in a certain algebra 2 class of 26 students, 10 of them play basketball and 5 of them play baseball. there…

in a certain algebra 2 class of 26 students, 10 of them play basketball and 5 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 26 students, 10 of them play basketball and 5 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:

$\frac{6}{13}$

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 $P(A\cap B)$

$P(A)=\frac{10}{26}$, $P(B)=\frac{5}{26}$, $P(A\cap B)=\frac{3}{26}$

Step3: Substitute values into the formula

$P(A\cup B)=\frac{10}{26}+\frac{5}{26}-\frac{3}{26}=\frac{10 + 5- 3}{26}=\frac{12}{26}=\frac{6}{13}$