Answer:
1. 45
2. 9
3. 11
4. 17
Step-by-step explanation:
The computation is shown below:
Given that
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.
Based on the above information
N(H)=28
N(C) 26
N(D) 14
N(H∪C) = 8
N(H∪D) = 4
[tex]N(C\cup D)=3[/tex]
[tex]N(H\cup C\cup D)=2[/tex]
We know that
[tex]Total=N(H)+N(C)+N(D)-N(H\cup C)-N(H\cup D)-N(C\cup D)+N(H\cup C\cup D)[/tex]
= 28 + 26 + 14 - 8 - 4 - 3 + 2
= 55
1. The students who didnt check any kind of the box is
= 100 - 55
= 45
2) Students who checked exactly two boxes is
[tex]= N(H\cup C)+N(H\cup D)+N(C\cup D)-3N(H\cup C\cup D)[/tex]
= 8 + 4 + 3 - 6
= 9
3) Students who checked a minimum of two boxes
[tex]= N(H\cup C)+N(H\cup D)+N(C\cup D)-2N(H\cup C\cup D)[/tex]
= 8 + 4 + 3 - 4
= 11
4) Given N(C) = 26
Now we have to deduct N(C∪D) and N(H∪C) as it could be checked aside from the club
= 26 - 8 - 3
= 15
Here the [tex]N(H\cup C\cup D)[/tex] could be deducted twice under both categories so it would be added one time to neutralize
= 15 + 2
= 17
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