Let two subsets $P,Q$ be selected from the set $A=\{1,2,3,4,5\}$. Find the probability... 
Let two subsets $P,Q$ be selected from the set $A=\{1,2,3,4,5\}$. Find the probability that



*

*A) $P\cap Q=\phi$

*B) $P\cup Q=A$

*C) $P\cap Q$ contains exactly one element


My Attempt:
For part A), each element has $3$ options. It can either go to $P$ or to $Q$ or to neither. So, total cases$=3^5$
Cases involving $P\cap Q$ can be: $^5C_0(^5C_0+...^5C_5)+^5C_1(^4C_0+...^4C_4)+^5C_2(^3C_0+...^3C_3)+^5C_3(^2C_0+...^2C_2)+^5C_4(^1C_0+^1C_1)+^5C_5(^5C_0)=2^5+5(2^4)+10(2^3)+10(2^2)+5(2)+1$
Is this correct? Is there an easier way to solve it?
For part $B),$ each element has just two options. So, total cases$=2^5$
So, favorable cases$=^5C_0(^5C_5)+^5C_1(^4C_4+^5C_5)+^5C_2(^3C_3+^4C_4+^5C_5)+...+^5C_5(^5C_0+^5C_1+...+^5C_5)$
Or, maybe $^5C_0(^5C_5)+^5C_1(^5C_4+^5C_5)+^5C_2(^5C_3+^5C_4+^5C_5)+...+^5C_5(^5C_0+^5C_1+...+^5C_5)$
Not sure which one to go with.
For part $C)$, maybe we can take $^5C_1$. This element would be in the intersection and for the remaining $4$ elements, we can approach like in part $A)$?
 A: Let $P, Q$ be labeled sets (to avoid fussy counting arguments).  Then our sample space consists of ordered pairs $(P, Q)$, where $P, Q \subseteq A = \{1, 2, 3, 4, 5\}$.  Since a set with $n$ elements has $2^n$ subsets, there are $2^5$ ways to select $P$ and $2^5$ ways to select $Q$.  Hence, there are
$$2^5 \cdot 2^5 = 2^{10}$$
elements in our sample space.

Find the probability that $P, Q \subseteq A$ satisfy $P \cap Q = \emptyset$.

If $P$ and $Q$ are disjoint, then each element of $A$ is in exactly one of the sets $P$, $Q$, or $(P \cup Q)'$.  Hence, there are $3^5$ favorable cases.  Thus,
$$\Pr(P \cap Q = \emptyset) = \frac{3^5}{2^{10}} = \frac{243}{1024}$$

Find the probability that $P, Q \subseteq A$ satisfy $P \cup Q = A$.

If $P \cup Q = A$, then each element of $A$ is in exactly one of the sets $P - Q, P \cap Q, Q - P$.  Hence, there are $3^5$ favorable cases.  Thus,
$$\Pr(P \cup Q = A) = \frac{3^5}{2^{10}} = \frac{243}{1024}$$

Find the probability that $P \cap Q$ contains exactly one element.

There are five choices for the element in $P \cap Q$.  Each of the other four elements must be in exactly one of the sets $P$, $Q$, $(P \cup Q)'$.  Hence, there are $5 \cdot 3^4$ favorable cases.  Hence,
$$\Pr(|P \cap Q| = 1) = \frac{5 \cdot 3^4}{2^{10}} = \frac{405}{1024}$$
A: A) $P$ is any subset of size $0,1,2,3,4$ or $5$, $Q$ is any subset of the complement of $P$. The total number of pairs $(P,Q)$:
$\sum_{k=0}^5 {5\choose k}2^{5-k}=(2+1)^5=243$. The probability is $\frac{243}{2^5\cdot 2^5}=\frac{243}{1024}$.
B) $P$ is any subset of size $0,1,2,3,4$ or $5$, $Q$ is any subset which is a union of the complement of $P$ and a subset of $P$. The number of pairs $(P,Q)$ is $\sum_{k=0}^5 {5\choose k}\cdot 2^k=243$.
The probability is $\frac{243}{2^5\cdot 2^5}=\frac{243}{1024}$.
C) The intersection element $x$ is $1,2,3,4,5$ (5 choices). The sets $P,Q$ are disjoint subsets of $\{1,2,3,4,5\}\setminus \{x\}$. See part A) with $5$ replaced by $4$. The number of pairs is $5\cdot 3^4=405$, the probability is $\frac{405}{1024}$.
