Solution to in a survey of 100 students, 56 wrote the Maths exams, 23 wrote psychology and … - Sikademy
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Archangel Macsika

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. Search Results Web results

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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

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Question ID: mtid-5-stid-8-sqid-4007-qpid-2706