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In a class of 50 students, 30 like Cricket, 20 like Football, and 10 like both. How many students like neither?
5
10
15
20
10
Subtract the intersection from the sum of individual sets to find the union (30+20-10=40), then subtract that result from the total (50-40=10).
Total students = 50, Students who like Cricket = 30, Students who like Football = 20, Students who like both = 10
n(AтИкB)=n(A)+n(B)тИТn(AтИйB)
Subtract the intersection from the sum of individual sets to find the union (30+20-10=40), then subtract that result from the total (50-40=10).
Students often add 30 and 20 to get 50, mistakenly assuming that all students like at least one, and fail to subtract the 10 students who are counted twice.
Define the Sets
Let C be the set of students who like Cricket and F be the set of students who like Football. Given n(C)=30, n(F)=20, and n(CтИйF)=10.
n(C)=30,n(F)=20,n(CтИйF)=10
Calculate Union of Sets
Find the number of students who like at least one sport using the Inclusion-Exclusion principle: n(CтИкF)=n(C)+n(F)тИТn(CтИйF).
n(CтИкF)=30+20тИТ10=40
Calculate Neither
Subtract the number of students who like at least one sport from the total number of students in the class.
50тИТ40=10
B is correct because 10 students like neither Cricket nor Football.
This logic is identical to solving problems involving probability of at least one event occurring, where you subtract the intersection of overlapping probabilities.