If P, Q, and R are sets such that; n(P) = 20, n(Q) = 16, n(R) = 21, n(P∩Q) = 7, n(P∩R)=8, n(Q∩R) = 5 and R(P∩Q∩R) = 3. a. represent this information on a Venn Diagram. b. Find; i. n(PUQUR) ii. the probability of (PUQ)' U R

We start with the innermost intersection and work outwards:

• $n(P \cap Q \cap R) = 3$
• $n(P \cap Q \text{ only}) = n(P \cap Q) - n(P \cap Q \cap R) = 7 - 3 = 4$
• $n(P \cap R \text{ only}) = n(P \cap R) - n(P \cap Q \cap R) = 8 - 3 = 5$
• $n(Q \cap R \text{ only}) = n(Q \cap R) - n(P \cap Q \cap R) = 5 - 3 = 2$
• $n(P \text{ only}) = n(P) - (n(P \cap Q \text{ only}) + n(P \cap R \text{ only}) + n(P \cap Q \cap R))$ $n(P \text{ only}) = 20 - (4 + 5 + 3) = 20 - 12 = 8$
• $n(Q \text{ only}) = n(Q) - (n(P \cap Q \text{ only}) + n(Q \cap R \text{ only}) + n(P \cap Q \cap R))$ $n(Q \text{ only}) = 16 - (4 + 2 + 3) = 16 - 9 = 7$
• $n(R \text{ only}) = n(R) - (n(P \cap R \text{ only}) + n(Q \cap R \text{ only}) + n(P \cap Q \cap R))$ $n(R \text{ only}) = 21 - (5 + 2 + 3) = 21 - 10 = 11$

Since I cannot draw a diagram, here are the values for each region:

• Elements in $P$ only: 8
• Elements in $Q$ only: 7
• Elements in $R$ only: 11
• Elements in $P \cap Q$ only (not in $R$): 4
• Elements in $P \cap R$ only (not in $Q$): 5
• Elements in $Q \cap R$ only (not in $P$): 2
• Elements in $P \cap Q \cap R$: 3

Using the Principle of Inclusion-Exclusion:

Substitute the given values:

The set $(P \cup Q)' \cup R$ consists of all elements that are either not in $P$ or $Q$, or are in $R$.
From the Venn Diagram regions, the elements not in $P$ or $Q$ are those in $R \text{ only}$ (which is 11), plus any elements outside $P \cup Q \cup R$.
The elements in $R$ are $n(R) = 21$.
The union $(P \cup Q)' \cup R$ can be understood as all elements in $R$, plus any elements that are outside $P \cup Q \cup R$.
If we assume there are no elements outside $P \cup Q \cup R$ (i.e., $n(\mathcal{U}) = n(P \cup Q \cup R)$), then the elements not in $P$ or $Q$ are simply the elements in $R \text{ only}$.
In this case, $(P \cup Q)' \cup R = (R \text{ only}) \cup R = R$.
So, $n((P \cup Q)' \cup R) = n(R) = 21$.

To calculate probability, we need a universal set $n(\mathcal{U})$. Since no universal set is specified, we assume the universal set is $P \cup Q \cup R$.
So, $n(\mathcal{U}) = n(P \cup Q \cup R) = 40$.
The probability is:

The probability of $(P \cup Q)' \cup R$ is $\boxed{\frac{21}{40}}$.