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 approach this problem using the formula:
P(A and B) = P(A) + P(B) - P(A or B)
where P(A) is the probability of event A, P(B) is the probability of event B, and P(A or B) is the probability of either event A or event B (or both) occurring.
In this case, let A be the event that a student plays basketball, and B be the event that a student plays baseball. We are given that P(A) = 11/23, P(B) = 13/23, and P(neither sport) = 8/23. We want to find P(A and B), the probability that a student plays both sports.
To use the formula, we need to first find P(A or B). This can be calculated as:
P(A or B) = P(A) + P(B) - P(A and B)
where P(A and B) is the probability that a student plays both basketball and baseball. We don't know this value yet, so we'll leave it as a variable.
Substituting in the given values, we get:
P(A or B) = (11/23) + (13/23) - P(A and B)
P(A or B) = 24/23 - P(A and B)
We know that the probability of either playing basketball or baseball (or both) is less than or equal to 1. So we can write:
P(A or B) ≤ 1
Substituting this inequality into the previous equation, we get:
24/23 - P(A and B) ≤ 1
P(A and B) ≥ -1/23 + 24/23
P(A and B) ≥ 23/23
P(A and B) ≥ 1
This tells us that the probability of a student playing both basketball and baseball is greater than or equal to 1 (which makes sense - you can't play both if you don't play either!). So our answer is:
P(A and B) ≥ 1
But since probabilities must be between 0 and 1, this means that:
P(A and B) = 1
So the probability that a student chosen randomly from the class plays both basketball and baseball is 1.