In a certain Algebra 2 class of 21 students, 6 of them play basketball and 10 of them play baseball. There are 7 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?
make a Venn diagram, label the intersection of both circles as x
So x students would play both sports.
It is easy to see that
6-x + x + 10-x + 7 = 21
-x = -2
x = 2
So 2 students play both sports,
prob(picking one of those 2) = 2/21
the answer is 4/21
To find the probability that a student chosen randomly from the class plays both basketball and baseball, we need to first determine the total number of students who play both sports.
Let's break down the information given:
- Total number of students in the class = 21
- Number of students who play basketball = 6
- Number of students who play baseball = 10
- Number of students who play neither sport = 7
We can visualize this information using a Venn diagram. Let's represent the sets as follows:
Let's denote:
- Number of students who play both basketball and baseball as "x"
- Number of students who play only basketball as "y"
- Number of students who play only baseball as "z"
Total number of students who play basketball = x + y
Total number of students who play baseball = x + z
Now, we can use the given information to determine the values of x, y, and z.
Since 6 students play basketball, and there are x students who play both basketball and baseball, the number of students who play only basketball (y) would be 6 - x.
Similarly, since 10 students play baseball, and there are x students who play both basketball and baseball, the number of students who play only baseball (z) would be 10 - x.
Now, let's calculate the value of x. We know that the total number of students in the class is 21 and the number of students who play neither sport is 7. So, the total number of students who play basketball or baseball is (21 - 7) = 14.
We can express the total number of students who play only basketball or only baseball using the values of y and z:
Total number of students who play only basketball or only baseball = y + z = (6 - x) + (10 - x) = 16 - 2x
Since the total number of students who play only basketball or only baseball is 16 - 2x, and the total number of students who play both basketball and baseball is x, we can set up the equation:
(16 - 2x) + x = 14
Simplifying the equation:
16 - x = 14
-x = 14 - 16
-x = -2
Multiplying both sides by -1:
x = 2
Therefore, there are 2 students who play both basketball and baseball.
To find the probability that a student chosen randomly from the class plays both sports, we divide the number of students who play both basketball and baseball (2) by the total number of students in the class (21):
Probability = Number of students playing both sports / Total number of students
Probability = 2 / 21
Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is 2/21 or approximately 0.0952.
please check for typos. I get 22 total
14 play something
5 basketball alone
9 baseball alone
1 both
that means 15 not 14 play something
I suspect 6 play neither or there are 22 students