Question

In a certain algebra 2 class of 30 students, 16 of them play basketball and 12 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?

Answer 1

We first explain the given numerical forms of the $n\left( A \right),n\left( B \right),n\left( A\cup B \right),n\left( A\cap B \right)$ where sets A and B are for students playing basketball and baseball respectively. We try to express the $n\left( A\cap B \right)$ with $n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)$. We put the values to find the probability for $p\left( A\cap B \right)$ using $p\left( A\cap B \right)=\dfrac{n\left( A\cap B \right)}{n\left( S \right)}$.

Complete step by step answer:
The given problem is the problem of probability in set inclusion.
We assume two sets A and B for students playing basketball and baseball respectively.
In total there are 30 students out of which 5 students play neither sport. This means $n\left( A\cup B \right)=30-5=25$. We take all students as set S where $n\left( S \right)=30$.
There are 16 students who play basketball and 12 play baseball.
So, $n\left( A \right)=16$ and $n\left( B \right)=12$.
We have to find the probability that a student chosen randomly from the class plays both basketball and baseball.
The number of students that play both basketball and baseball can be denoted as $n\left( A\cap B \right)$. We know that $n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)$.
Placing the values, we get $n\left( A\cap B \right)=16+12-25=3$.
We know that the fundamental theorem of probability gives us that $p\left( A\cap B \right)=\dfrac{n\left( A\cap B \right)}{n\left( S \right)}$.
So, $p\left( A\cap B \right)=\dfrac{3}{30}=\dfrac{1}{10}$. The probability of students chosen randomly from the class who play both basketball and baseball is $\dfrac{1}{10}$.

Note: We need to remember that the relation between numerical values and their probabilities is similar for all the given $n\left( A \right),n\left( B \right),n\left( A\cup B \right),n\left( A\cap B \right)$. That’s why we didn’t use the concept of number of points in a set and instead we directly used the numerical form to find the solution. To find individual probabilities, we divide them with $n\left( S \right)$.