Question

find n(AuBuC), if n (A) =16 n (B) =12 ,n(C)=14,n(AuB) =6, n(BnC)=7,n(AnC)=5 and n(AnBnC)=20

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Answer 1

Answer:

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Step-by-step explanation:

n(AuBuC) = n(A)+n(B)+ n(C)-n(AnB)-n(Anc)-n(BnC)+n(AnBnC)

As we know n(AuB)=n(A)+ n(B)-n(AnB)

Here when we add A and B we are adding (AnB) two times , as it is present in both A and B. So we subtract (AnB) one time from A and B.

Which means we need to take the area of A and B set but no area should repeat.

When we take three sets A,B and C

We take area of A+ Area of B + area of C ————eq

But they are intersecting so the area we are taking is greater than actual area. We have again added (AnB), (AnC) and (BnC) twice. Therefore we subtract the intersecting area from eq one time as done above

A+B+C-(AnB)-(AnC)-(BnC)

But when we do this, area of (AnBnC) is also removed as all A,B,C,(Anb),(Anc),(BnC) all of them contain (AnBnC) in them.

Therefore we add a (AnBnC) in eq.

What is the meaning of P (AUBUC)?

What is the proof of the formula n(AUBUC)?

What is N (AUBUCU) =?

What is the value of (A U B U C) ∩ (A ∩ B' ∩ C') ∩ C'?

How do you prove that P (AuBuC) =1-p(A') p(B') p(C')?