please answer the question
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please answer the question
write full solution
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Answer:
1 is the correct answer I guess
Answer:
Option 3 is the answer
Step-by-step explanation:
Total number of students=60
No.of students who do not play any game =15
No.of students who play games =60-15=45
No.of students who play hockey =n(H)=23
No.of students who play basketball=n(B)=15
No.of students who play cricket=n(C)=20
No.of students who paly both hockey and basketball=n(HnB)=7
No.of students who play both basketball and Cricket=n(BnC)=5
No.of students who play both Hockey and Cricket=n(HnC)=4
Let no.of students who play all the three games be X =n(HnBnC)=X
We know that
n(HUBUC)=n(H)+n(B)+n(C)- n(HnB)- n(BnC)- n(HnC) +n(HnBnC)
45=23+15+20-7-5-4+X
=>45=58-16+X
=>45=42+X
=>45-42=X
=>X=3
No.of students who play all the three games =3
No.of students who play only hockey
=23-4-1-3=15
No.of students who play hockey and cricket but not Basketball = n(HnC)-n(HnBnC)=4-3=1