The answer is ten. The formula given earlier contained an error.
Removing the 18 individuals that don't play from the 50 surveyed leaves 32 in the union of the three circular regions.
I have labeled them H, C, and F. In this tri-ring area, those who play Hockey "or" Cricket "or" Football are represented.
If I use the G notation described above, the formula for this number of members in this union is:
g(HuCuF)= g(H)+g(C)+g(F)-g(H&C)-g(H&F)-g(C&F)+g(H&C&F)
inputing our values gives:
....
32 = 20 + 15 + 11 - 7 - 5 - 4 + x ; where
x is the innermost intersection (region 5 in my manner of labeling)
Simplifying now:
32 = 46 - 16 + x
32 = 30 + x
2 = x
NOTE: The error before is that we do not add double region 5. The formula only calls for it to be added once.
It was included 3 times with H, C, And F and then removed 3 times as we subtracted the two way intersections
(which the three way is part of) and so must be put back again once for inclusion, but not twice.
With
x=2 we know that region 2 (players of H&C but not F) has 7-2=5 members
that region 4 (players of H&F but not C) has 5-2=3 members and
that region 6 (players of C&F but not H) has 4-2=2 members
So the sum of the members of Regions 2, 4, & 6 are 5+3+2=10
NOTE: A set up my venn and label the regions as shown here:
Here are the results of my calculations.
Once I have the two. I fillin from the innermost region outwards,
subtracting as I go what I have already included so far.
In the end What's inside adds to 32 and meets the other conditions required.
I don't think I missed anything. I hope this is a helpful resolution.
