Problem:
In a class of 95 students, each student plays at least one of hockey, cricket, and football. If 55 play hockey, 60 play cricket and 65 play football, the number of students who play all the three games could be at most Fa? Explain in details.

Solution:
To solve the problem, we need to use the principle of inclusion-exclusion.

Step 1: Find the total number of students who play at least one game.
Total = hockey + cricket + football - (hockey and cricket) - (cricket and football) - (football and hockey) + (hockey, cricket, and football)
Total = 55 + 60 + 65 - (hockey and cricket) - (cricket and football) - (football and hockey) + (hockey, cricket, and football)
Total = 180 - (hockey and cricket) - (cricket and football) - (football and hockey) + (hockey, cricket, and football)

Step 2: Find the maximum number of students who could be playing all three games.
We know that the total number of students is 95. Therefore,
95 = 180 - (hockey and cricket) - (cricket and football) - (football and hockey) + (hockey, cricket, and football)
(hockey and cricket) + (cricket and football) + (football and hockey) - (hockey, cricket, and football) = 85
2(hockey, cricket, and football) - (hockey and cricket) - (cricket and football) - (football and hockey) = 85
2(hockey, cricket, and football) = 85 + (hockey and cricket) + (cricket and football) + (football and hockey)
Since the maximum value of any intersection is less than or equal to the smallest set, the maximum value of (hockey and cricket), (cricket and football), and (football and hockey) is 55, which is the size of the smallest set (hockey). Therefore,
2(hockey, cricket, and football) ≤ 85 + 55 + 55 + 55
2(hockey, cricket, and football) ≤ 250
hockey, cricket, and football ≤ 125

Step 3: Conclusion
The number of students who play all three games could be at most 125.