Unknownnn's question from Yahoo Answers (re: finite math/set theory)

[Chris L T521]

Here is the question:

Let U be a universal set with subsets A and B such that n(U) = 130, n(A′) = 55, n(B′) = 69, and n(A∩B) = 23. Find n(A∪B).

n(A∪B) = halp? plox

Here is a link to the question:

Finite math problem involving venn diagrams? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Chris L T521]

Hi Unknownnn,

We will use the following identities/equations in our computation:

- $n(A^{\prime})=n(U)-n(A)$;
- $n(A\cup B)=n(A)+n(B)-n(A\cap B)$
- $n((A\cap B)^{\prime})=n(A^{\prime}\cup B^{\prime})$ and $n((A\cup B)^{\prime})=n(A^{\prime}\cap B^{\prime})$ (De Morgan's Laws)

To get the answer we seek, let us use the second equation, but with complements instead of regular sets since we know the values of $n(A^{\prime})$ and $n(B^{\prime})$:
\[n(A^{\prime}\cup B^{\prime})=n(A^{\prime})+n(B^{\prime})-n(A^{\prime}\cap B^{\prime}).\]
Using the equations I provided above, we see that
\[\begin{aligned}n(A^{\prime}\cup B^{\prime})=n((A\cap B)^{\prime})= n(U)-n(A\cap B) &= n(A^{\prime})+n(B^{\prime}) -n((A\cup B)^{\prime})\\ &= n(A^{\prime})+n(B^{\prime}) -(n(U)-n(A\cup B))\\ &= n(A^{\prime})+n(B^{\prime})-n(U)+n(A\cup B)\end{aligned} \]

Solving for $n(A\cup B)$ gives us
\[n(A\cup B)=2n(U)-n(A^{\prime})-n(B^{\prime})-n(A\cap B)=260-55-69-23=113\]

Thus, $n(A\cup B)=113$.