In a certain Algebra 2 class of 26 students, 21 of them play basketball and 7 of them play baseball. There are 2 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

People who play baseball AND basketball:

4
so the answer is 4/26 or if u wanna simplify, 2/13

In a certain Algebra 2 class of 26 students, 18 of them play basketball and 7 of them play baseball. There are 5 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?

Oh, the probability of a student playing both basketball and baseball in that Algebra 2 class is like finding a unicorn at a clown convention – pretty rare! Let's do some calculations to figure it out.

First, we have 26 students in total, and we know that 21 of them play basketball and 7 play baseball. Since there are 2 students who play neither sport, we can subtract them from the total number of students involved in sports, which is 21 + 7 = 28.

Now, we want to find the probability of a student playing both basketball and baseball. We can use the formula:

Probability = Number of favorable outcomes / Total number of outcomes

But we need to figure out the favorable outcome first. Since we have subtracted the 2 students who play neither sport, our favorable outcome becomes 28 - 26 = 2.

Now, the total number of outcomes is the total number of students, which is 26.

Plugging these values into the formula, the probability is:

Probability = 2 (favorable outcomes) / 26 (total outcomes)

Simplifying, we get:

Probability = 1 / 13

So, the probability that a student chosen randomly from the class plays both basketball and baseball is 1/13. That's almost as rare as finding a clown without a red nose!

To find the probability that a student plays both basketball and baseball, we need to determine the number of students who play both sports and divide it by the total number of students in the class.

Given that there are 26 students in total in the class, and 2 students play neither sport, we can deduce that 26 - 2 = 24 students play at least one of the two sports.

Let's set up a Venn diagram to help visualize this:

_________
| |
Basketball | |
(21) | A |
| |
|_________|
(x)
| |
Baseball | B |
(7) | |
|_________|

Here, A represents the set of students who play basketball (21 students), B represents the set of students who play baseball (7 students), and x represents the intersection (students who play both sports).

Since we are looking for the probability that a student plays both basketball and baseball, we need to find the value of x.

We know that the total number of students who play at least one of the two sports is 24. Thus, we can use the principle of inclusion-exclusion to find x:

x = (Number of students who play basketball) + (Number of students who play baseball) - (Number of students who play neither sport)
x = 21 + 7 - 2
x = 26

Therefore, 26 students play both basketball and baseball.

Now, to find the probability of randomly selecting a student who plays both sports:

Probability = (Number of students who play both sports) / (Total number of students)
Probability = 26 / 26
Probability = 1

So, the probability that a student chosen randomly from the class plays both basketball and baseball is 1, or 100%.

24 play something, so if x play both,

21+7-x = 24
and P(both) = x/24