Let's define some variables to organize the information:

• Let $n = 28$ be the total number of students in the class.
• Let $B_b$ be the number of students who play basketball, which is $B_b = 21$.
• Let $B_s$ be the number of students who play baseball, which is $B_s = 20$.
• Let $N$ be the number of students who play neither sport, which is $N = 5$.
• Let $x$ be the number of students who play both basketball and baseball, which is what we need to find.

Step 1: Find the number of students who play either basketball or baseball or both.

Since 5 students play neither sport, the number of students who play either basketball or baseball or both is: $28 - 5 = 23$ So, 23 students are involved in either or both sports.

Step 2: Use the inclusion-exclusion principle to find the number of students who play both sports.

The inclusion-exclusion principle states that: $\text{(Students who play either sport)} = B_b + B_s - \text{(Students who play both sports)}$ Substitute the known values: $23 = 21 + 20 - x$ Now solve for $x$: $23 = 41 - x$ $x = 41 - 23 = 18$

Step 3: Calculate the probability.

The probability that a randomly chosen student plays both basketball and baseball is the ratio of the number of students who play both sports to the total number of students: $P(\text{Both sports}) = \frac{x}{n} = \frac{18}{28} = \frac{9}{14}$

Thus, the probability that a student chosen randomly plays both basketball and baseball is $\frac{9}{14}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the probability that a randomly chosen student plays only basketball?
2. What is the probability that a randomly chosen student plays only baseball?
3. How many students play only basketball?
4. How many students play only baseball?
5. What is the probability that a student plays either basketball or baseball but not both?

Tip: The inclusion-exclusion principle is helpful for solving problems involving overlapping groups!