Answer:
Cricket only = 30
Volleyball only = 15
Hockey only = 25
Explanation:
The number of cricket players is represented as n(C).
The count of hockey players as n(H).
The total volleyball participants as n(V).
The available data indicates;
n(C) = 50, n(H) = 50, n(V) = 40
For students playing both cricket and hockey, we denote as n(C∩H).
Participants in both hockey and volleyball as n(H∩V).
Players in both cricket and volleyball as n(C∩V).
Students engaging in all three sports as n(C∩H∩V).
From the data given;
n(C∩H) = 15
n(H∩V) = 20
n(C∩V) = 15
n(C∩H∩V) = 10
Thus, the total number of students who play at least one sport:
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
Consequently, the overall total number of students, n(U) = 100.
Note: n(U) represents the universal set.
Let a = number of individuals who only played cricket and volleyball.
Let b = number of individuals who only played cricket and hockey.
Let c = number of individuals who only participated in hockey and volleyball.
Let d = number who played all three sports.
This allows us to establish:
d = n(CnHnV) = 10
n(CnV) = a + d = 15
n(CnH) = b + d = 15
n(HnV) = c + d = 20
From here, we deduce:
a = 15 - 10 = 5
b = 15 - 10 = 5
c = 20 - 10 = 10
Thus, for the cricket-only player count;
n(C) - [a + b + d] = 50 - (5 + 5 + 10) = 30
For hockey-only players:
n(H) - [b + c + d] = 50 - (5 + 10 + 10) = 25
And for volleyball-only participants:
n(V) - [a + c + d] = 40 - (10 + 5 + 10) = 15
