Let $A, B,C$ be subsets of $\{1,2,...n\}$. What is $P((A\cap B)\subseteq C)$? My reckoning is that there are totally $2^{n} \cdot 2^{n} \cdot2^{n}$ options for $3$ subsets and for $(A\cap B)\subseteq C$ ,  for each of $\sum_{k=0}^{n}\binom{n}{k}$ options for C there are $\sum_{i=0}^{k}\binom{k}{i}$ for $A$ and $\sum_{i=0}^{k}\binom{k}{i}$ for $B$ and so in total we have:
$$P((A\cap B)\subseteq C)=\frac{\sum_{k=0}^{n}\binom{n}{k}\cdot 2^{k}\cdot 2^{k}}{2^{n}\cdot 2^{n}\cdot 2^{n}}=\frac{\sum_{k=0}^{n}\binom{n}{k}\cdot 4^{k}}{8^{n}}=\frac{5^{n}}{8^{n}}=(\frac{5}{8})^{n}$$
Is this reckoning right?
 A: We want to calculate $P(A\cap B\subseteq C)=P((A\cap B)\setminus C=\emptyset)$.
For each $i$ in $\{1,2,\dots,n\}$ let $E_i$ be the event $i\notin(A\cap B)\setminus C$.
As the events $E_1,E_2,\dots,E_n$ are independent, we have
$$P((A\cap B)\setminus C=\emptyset)=P(E_1\cap E_2\cap\cdots\cap E_n)=P(E_1)P(E_2)\cdots P(E_n)=\left(\frac78\right)^n$$
since
$$P(E_i)=1-P(i\in(A\cap B)\setminus C)=1-P(i\in A,\ i\in B,\ i\notin C)=1-\frac18=\frac78.$$
A: Along the lines of your attempt, here's a corrected version . . .

Fix a positive integer $n$ and let $X=\{1,...,n\}$.

Suppose $A,B,C$ are subsets of $X$, chosen independently at random from a uniform distribution on the power set of $X$.

For each $k\in\{0,...,n\}$ there are ${\large{\binom{n}{k}}}$ choices for $C$ with $|C|=k$.

Fix some such choice of $k$ and $C$.

For each $j\in\{0,...,k\}$ there are are ${\large{\binom{k}{j}}}$ choices for $A\cap B$ with $(A\cap B)\subseteq C$ and $|A\cap B|=j$.

Fix some such choice of $j$ and $A\cap B$.

Each of the $n-j$ elements of $X\,{\large{\setminus}}(A\cap B)$ can be assigned to

*

*$A$ but not $B$.$\\[4pt]$

*$B$ but not $A$.$\\[4pt]$

*neither of $A,B$.

Thus for the fixed choice of $j$ and $A\cap B$, there are $3^{n-j}$ choices for $A,B$.

Then $P\bigl((A\cap B)\subseteq C\bigl)=\dfrac{w}{8^n}$, where
\begin{align*}
w
&=
\sum_{k=0}^n
\binom{n}{k}
\sum_{j=0}^k
\binom{k}{j}3^{n-j}
\\[4pt]
&=
3^n\sum_{k=0}^n
\binom{n}{k}
\sum_{j=0}^k
\binom{k}{j}\left(\frac{1}{3}\right)^j
\\[4pt]
&=
3^n\sum_{k=0}^n
\binom{n}{k}\left(\frac{4}{3}\right)^k
\\[4pt]
&=
3^n\left(\frac{7}{3}\right)^n
\\[4pt]
&=
7^n
\end{align*}
