There are a couple of ways to approach this problem, but one common method is to use the formula:
P(A or B) = P(A) + P(B) - P(A and B)
where P(A) is the probability of event A, P(B) is the probability of event B, and P(A and B) is the probability of both events happening together.
Using this formula, we can first find the probability of a random student playing basketball:
P(basketball) = 14/28 = 1/2
Similarly, we can find the probability of a random student playing baseball:
P(baseball) = 10/28 = 5/14
To find the probability of a student playing either basketball or baseball, we just need to add these probabilities together:
P(basketball or baseball) = P(basketball) + P(baseball) - P(basketball and baseball)
To find the probability of both basketball and baseball, we need to subtract the probability of neither sport from the total number of students:
P(neither) = 5/28
Total students = 28
P(basketball and baseball) = Total students - P(neither) - P(basketball or baseball)
P(basketball and baseball) = 28 - 5 - (14+10-5)
P(basketball and baseball) = 8
Therefore,
P(basketball or baseball) = 1/2 + 5/14 - 8/28
P(basketball or baseball) = 0.5 + 0.357 - 0.286
P(basketball or baseball) = 0.571
So the probability of a student chosen randomly from the class playing basketball or baseball is 0.571, or approximately 57.1%.
In a certain Algebra 2 class of 28 students, 14 of them play basketball and 10 of them play are all. There are 5 students who play neither sport. What is the probability that a student chosen randomly from the class plays basketball or baseball?
1 answer