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?
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%.
To find the probability that a student chosen randomly from the class plays basketball or baseball, we can use the formula:
P(A or B) = P(A) + P(B) - P(A and B)
Where:
P(A or B) represents the probability of event A or event B happening,
P(A) represents the probability of event A happening,
P(B) represents the probability of event B happening,
P(A and B) represents the probability of both event A and event B happening.
In this case,
Event A: Playing basketball
Event B: Playing baseball
Given:
Total number of students in the class (n) = 28
Number of students who play basketball (A) = 14
Number of students who play baseball (B) = 10
Number of students who play neither sport (A and B') = 5
Let's calculate the probability using the formula:
P(A or B) = P(A) + P(B) - P(A and B)
P(A) = number of students who play basketball / total number of students = A / n
P(B) = number of students who play baseball / total number of students = B / n
P(A and B) = number of students who play both basketball and baseball / total number of students = (A and B) / n
Substituting the values, we get:
P(A or B) = (A / n) + (B / n) - ((A and B) / n)
P(A or B) = (14 / 28) + (10 / 28) - (5 / 28)
Simplifying the expression:
P(A or B) = 0.5 + 0.357 - 0.179
P(A or B) = 0.678
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is approximately 0.678 or 67.8%.