A group of 87 students play hockey, football, or volleyball. Given the number of students who play each sport and the number who play pairs of sports, find how many play all three sports and represent

$$87 = 129 - 53 + n(H \cap F \cap V)$$ $$87 = 76 + n(H \cap F \cap V)$$

$$n(H \cap F \cap V) = 87 - 76$$ $$n(H \cap F \cap V) = \boxed{11}$$

So, 11 students play all three games.

2. Using a Venn diagram, represent the information.
Let $x = n(H \cap F \cap V) = 11$.
The number of students playing exactly two games are:

• Only Hockey and Football: $n(H \cap F) - x = 21 - 11 = 10$
• Only Hockey and Volleyball: $n(H \cap V) - x = 15 - 11 = 4$
• Only Football and Volleyball: $n(F \cap V) - x = 17 - 11 = 6$

The number of students playing exactly one game are:

• Only Hockey: $n(H) - (10 + 4 + 11) = 40 - 25 = 15$
• Only Football: $n(F) - (10 + 6 + 11) = 42 - 27 = 15$
• Only Volleyball: $n(V) - (4 + 6 + 11) = 47 - 21 = 26$

A Venn diagram would show three overlapping circles for H, F, and V. The regions would contain the following numbers:

• Center (H $\cap$ F $\cap$ V): 11
• H only F: 10
• H only V: 4
• F only V: 6
• H only: 15
• F only: 15
• V only: 26

3. How many students play only 2 of the games?

$$n(\text{only 2 games}) = 10 + 4 + 6$$ $$n(\text{only 2 games}) = \boxed{20}$$

So, 20 students play only 2 of the games.

4. If a student is picked at random from the class, find the probability that the student plays only one of the games.

$$n(\text{only 1 game}) = 15 + 15 + 26$$ $$n(\text{only 1 game}) = 56$$

$$P(\text{only 1 game}) = \frac{\text{Number of students playing only 1 game}}{\text{Total number of students}}$$ $$P(\text{only 1 game}) = \frac{56}{87}$$

The probability is $\boxed{\frac{56}{87}}$.