Answer:
1. 45
2. 9
3. 11
4. 17
Explanation:
Given the following:
28 checked H
26 checked C
14 checked D
8 checked H and C
4 checked H and D
3 checked C and D
2 checked all.
Hence N(H) = 28
N(C) = 26
N(D) = 14
N(H U C) = 8
N(H U D) = 4
N(C U D) = 3
N(H U C U D) = 2
We also know that
Total = N(H) + N(C) + N(D) - N(H U C) - N(H U D) - N(C U D) + N(H U C U D)
Substituting the given values, we obtain
Total = 55
1. Students that didn't check any box = 100 - 55 = 45 students
2. Students who checked exactly two box
= N(H U C) + N(H U D) + N(C U D) - 3N(H U C U D) (from probability theorem)
Substituting the values, we have 8 + 4 + 3 - 6 = 9 students
3. Students who checked atleast two box =
The people who have checked all three are needed to be calculated once. Earlier, we subtracted them thrice so we add one time
N(H U C) + N(H U D) + N(C U D) - 2N(H U C U D) = 8 + 4 + 3 - 4 = 11 students
4. Given N(C) = 26
We subtract N(CUD) and N(HUC) as they have checked another apart from club.
26 -8 - 3 = 15
Now we could notice we have subtracted N(HUCUD) twice in both categories, so we add one time to neutralise
15 + 2 = 17
Hence N(only C) = 17 students.
