in a certain algebra 2 class of 21 students, 10 of them play basketball and 8 of them play baseball. there…

in a certain algebra 2 class of 21 students, 10 of them play basketball and 8 of them play baseball. there are 6 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 21 students, 10 of them play basketball and 8 of them play baseball. there are 6 students who play both sports. what is the probability that a student chosen randomly from the class plays basketball or baseball?

Answer

Explanation:

Step1: Use the formula for the union of two - sets

The formula for (P(A\cup B)) is (P(A\cup B)=P(A)+P(B)-P(A\cap B)). Here, (n(A) = 10) (number of basketball - players), (n(B)=8) (number of baseball - players), (n(A\cap B) = 6) (number of students who play both), and (n(S)=21) (total number of students).

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

(P(A)=\frac{n(A)}{n(S)}=\frac{10}{21}), (P(B)=\frac{n(B)}{n(S)}=\frac{8}{21}), (P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{6}{21}).

Step3: Calculate (P(A\cup B))

(P(A\cup B)=\frac{10}{21}+\frac{8}{21}-\frac{6}{21}=\frac{10 + 8-6}{21}=\frac{12}{21}=\frac{4}{7}).

Answer:

(\frac{4}{7})