In a certain Algebra 2 class of 23 students, 11 of them play basketball and 13 of them play baseball. There are 8 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?
We can start by using the formula:
P(A and B) = P(A) + P(B) - P(A or B)
where A represents the event of playing basketball, B represents the event of playing baseball, and P(A or B) represents the probability of playing either basketball or baseball or both.
We know that P(A) = 11/23 (11 students play basketball out of 23 total students) and P(B) = 13/23 (13 students play baseball out of 23 total students). We can also find P(A or B) by using the fact that there are 8 students who play neither sport, which means there are 23 - 8 = 15 students who play at least one of the sports. Therefore:
P(A or B) = 15/23
Now we can plug in these values into the formula:
P(A and B) = P(A) + P(B) - P(A or B)
P(A and B) = 11/23 + 13/23 - 15/23
P(A and B) = 9/23
So the probability that a student chosen randomly from the class plays both basketball and baseball is 9/23.