Question

in a survey of 100 students, 56 wrote the Maths exams, 23 wrote psychology and 21 wrote the science exam. 12 wrote both maths and psychology exams, 9 write the maths and science exams and 6 wrote both psychology and science exams. 5 students wrote neither. determine how many students wrote all three exams.

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Answer 1

We use the theory of inclusion and exclusion here.

Let $A,B,C$ denote the events of writing the Mathematics, Psychology and Science Exams respectively.

Let $U$ denote the universal set, so $N(U)=100$ .

Now, we have the following information:

$N(A)=56\\ N(B)=23\\ N(C)=21\\ N(A\cap B)=12\\ N(A\cap C)=9\\ N(B\cap C)=6\\ N(\bar{A}\cap\bar{B}\cap\bar{C})=5$

From the last one, using the fact that $\bar{A}\cap\bar{B}\cap\bar{C}=(A\cup B\cup C)^C$ , we have $N(A\cup B\cup C)=95$

We need $N(A\cap B\cap C)$

Using inclusion-exclusion principle, we have

$N(A\cup B\cup C)=N(A)+N(B)+N(C)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;-N(A\cap B)-N(A\cap C)-N(B\cap C)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+N(A\cap B\cap C)\\ \implies N(A\cap B\cap C)=N(A\cup B\cup C)-(N(A)+N(B)+N(C))-(-N(A\cap B)-N(A\cap C)-N(B\cap C))\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\ =95-56-23-21+12+9+6=22$