Question

How to get the answers?​

Answer 1

By inclusion/exclusion,

`n(P' \cup Q) = n(P') + n(Q) - n(P' \cap Q)`

We have

`n(\xi) = n(P) + n(P') \implies n(P') = 28 - n(P)`

so that

`n(P'  \cup Q) = (28 - n(P)) + n(Q) - 2n(P) = 28 - 3n(P) + n(Q)`

Now,

`n(\xi) = 28 \implies n(P \cup Q) = 28 - 7 = 21`

and by inclusion/exclusion,

`n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)`

Decompose
`Q` into the union of two disjoint sets:

`Q = (P \cap Q) \cup (P' \cap Q)`

Since they're disjoint,

`n(Q) = n(P\cap Q) + n(P'\cap Q) \implies n(Q) = n(P\cap Q) + 2n(P)`

`\implies n(P \cup Q) = n(P) + (n(P\cap Q) + 2n(P)) - n(P \cap Q)`

`\implies 21 = 3n(P)`

`\implies n(P) = 7`

From the Venn diagram, we see there are 3 elements unique to
`P` - by the way, this is the set
`P \cap Q'` - so
`n(P\cap Q) = 7-3 = 4`, and it follows that

`n(Q) = n(P\cap Q) + 2n(P) = 4 + 2*7 = 18`

Finally, we get for (a)

`n(P' \cup Q) = 28 - 3n(P) + n(Q) = 28 - 3*7 + 18 = \boxed{25}`

For (b), we have by inclusion/exclusion that

`n(P \cup Q') = n(P) + n(Q) - n(P \cap Q') = 7 + 18 - 3 = \boxed{22}`