Out of a class of 100 students, 25 play at least cricket and football,

Out of a class of 100 students, 25 play at least cricket and football, 15 play at least cricket and hockey, 12 play at least football and hockey and 10 play all the three sports. The number of students playing cricket, football and hockey are 50, 37 and 22, respectively. The number of students who do NOT play any of the three sports is

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UPSC CAPF – 2024

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The correct option is A.

Let C, F, and H represent the sets of students who play Cricket, Football, and Hockey, respectively.
Total students = 100.
Given:
|C| = 50
|F| = 37
|H| = 22
|C ∩ F| = 25 (students playing at least cricket and football implies the intersection)
|C ∩ H| = 15 (students playing at least cricket and hockey implies the intersection)
|F ∩ H| = 12 (students playing at least football and hockey implies the intersection)
|C ∩ F ∩ H| = 10 (students playing all three sports)

We need to find the number of students who do NOT play any of the three sports. This is given by Total students – |C ∪ F ∪ H|.
We use the Principle of Inclusion-Exclusion for three sets:
|C ∪ F ∪ H| = |C| + |F| + |H| – (|C ∩ F| + |C ∩ H| + |F ∩ H|) + |C ∩ F ∩ H|
Substitute the given values:
|C ∪ F ∪ H| = 50 + 37 + 22 – (25 + 15 + 12) + 10
|C ∪ F ∪ H| = 109 – (52) + 10
|C ∪ F ∪ H| = 109 – 52 + 10
|C ∪ F ∪ H| = 57 + 10 = 67.

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The number of students who play at least one sport is 67.
The number of students who do NOT play any of the three sports = Total students – |C ∪ F ∪ H|
Number of students who do not play any sport = 100 – 67 = 33.

The phrasing “at least cricket and football” sometimes can be ambiguous. In standard set theory problems like this, it usually refers to the size of the intersection C ∩ F. If it meant only those who play exactly cricket and football (and not hockey), the numbers would be derived differently (e.g., |C ∩ F| – |C ∩ F ∩ H| would be the number playing *only* C and F). However, given the standard structure of such problems and the options, the interpretation used (intersection size) is almost certainly the intended one.
We can also visualize this with a Venn diagram. The value 10 goes in the center. The values for exactly two sports are:
Only C and F = |C ∩ F| – |C ∩ F ∩ H| = 25 – 10 = 15
Only C and H = |C ∩ H| – |C ∩ F ∩ H| = 15 – 10 = 5
Only F and H = |F ∩ H| – |C ∩ F ∩ H| = 12 – 10 = 2
The values for exactly one sport are:
Only C = |C| – (Only C&F) – (Only C&H) – (C&F&H) = 50 – 15 – 5 – 10 = 20
Only F = |F| – (Only C&F) – (Only F&H) – (C&F&H) = 37 – 15 – 2 – 10 = 10
Only H = |H| – (Only C&H) – (Only F&H) – (C&F&H) = 22 – 5 – 2 – 10 = 5
Total playing at least one sport = (Only C) + (Only F) + (Only H) + (Only C&F) + (Only C&H) + (Only F&H) + (C&F&H)
= 20 + 10 + 5 + 15 + 5 + 2 + 10 = 67.
Number not playing any sport = 100 – 67 = 33. This confirms the result.

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