100 students were asked to fill out a form with three survey questions, as follows: h: honor roll c: club membership (robotics club or gaming club) d: double-major survey results were as follows: 28 checked h (possibly non-exclusively), 26 checked c (possibly non-exclusively), 14 checked d (possibly non-exclusively) 8 checked h and c (possibly. non-exclusively), 4 checked h and d (possibly. non- exclusively), 3 checked c and d (possibly. non-exclusively) and 2 checked all three statements. 1. how many students didn't check any of the boxes? [a] 2. how many students checked exactly two boxes? [b] 3. how many students checked at least two boxes? [c]4. how many students checked the clubs box only?

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.