Answer:
Cricket only= 30
Volleyball only = 15
Hockey only = 25
Explanation:
Number of students that play cricket= n(C)
Number of students that play hockey= n(H)
Number of students that play volleyball = n(V)
From the question, we have that;
n(C) = 50, n(H) = 50, n(V) = 40
Number of students that play cricket and hockey= n(C∩H)
Number of students that play hockey and volleyball= n(H∩V)
Number of students that play cricket and volleyball = n(C∩V)
Number of students that play all three games= n(C∩H∩V)
From the question; we have,
n(C∩H) = 15
n(H∩V) = 20
n(C∩V) = 15
n(C∩H∩V) = 10
Therefore, number of students that play at least one game
n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)
= 50 + 50 + 40 – 15 – 20 – 15 + 10
Thus, total number of students n(U)= 100.
Note;n(U)= the universal set
Let a = number of people who played cricket and volleyball only.
Let b = number of people who played cricket and hockey only.
Let c = number of people who played hockey and volleyball only.
Let d = number of people who played all three games.
This implies that,
d = n (CnHnV) = 10
n(CnV) = a + d = 15
n(CnH) = b + d = 15
n(HnV) = c + d = 20
Hence,
a = 15 – 10 = 5
b = 15 – 10 = 5
c = 20 – 10 = 10
Therefore;
For number of students that play cricket only;
n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30
For number of students that play hockey only
n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25
For number of students that play volleyball only
n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15
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