1. Initial setup

We have: $E_1=\{1,2,3\},\quad F_1=\{1,3,4\},\quad G_1=\{2,3,4,5\}.$

Two elements are chosen from $E_1$ to form $S_1$. Since $E_1$ has 3 elements, the possible 2-element subsets are: $\{1,2\},\ \{1,3\},\ \{2,3\}$ each with probability $\frac13$.

Then: $E_2=E_1-S_1, \qquad F_2=F_1\cup S_1.$

Next, two elements are chosen from $F_2$ to form $S_2$. Then: $G_2=G_1\cup S_2.$

Finally, two elements are chosen from $G_2$ to form $S_3$, and $E_3=E_2\cup S_3.$

We are given the event: $E_3=E_1=\{1,2,3\}.$ We must find $p=P(S_1=\{1,2\}\mid E_3=E_1).$

Using conditional probability, $p=\frac{P(S_1=\{1,2\}\cap E_3=E_1)}{P(E_3=E_1)}.$

So we compute the probability of $E_3=E_1$ for each possible $S_1$.

2. Case 1: $S_1=\{1,2\}$

Then: $E_2=\{3\}, \qquad F_2=F_1\cup S_1=\{1,2,3,4\}.$

Now $S_2$ is any 2-element subset of $\{1,2,3,4\}$, so there are $\binom42=6$ possible choices, all equally likely.

Also, $G_2=G_1\cup S_2.$ Since $G_1=\{2,3,4,5\}$, the only new element $S_2$ can add is 1. Hence:

• if $1\in S_2$, then $G_2=\{1,2,3,4,5\}$;
• if $1\notin S_2$, then $G_2=\{2,3,4,5\}$.

We need $E_3=E_2\cup S_3=\{3\}\cup S_3=\{1,2,3\}.$ So necessarily, $S_3=\{1,2\}.$

Thus this is possible only if $1\in G_2$, i.e. only if $1\in S_2$.

Among the 6 possible $S_2$ choices, those containing 1 are: $\{1,2\},\{1,3\},\{1,4\},$ so 3 choices.

For each such $S_2$, we have $G_2=\{1,2,3,4,5\}$, and $S_3$ is chosen uniformly from its 2-element subsets. Number of such subsets: $\binom52=10.$ Only one favorable choice gives $S_3=\{1,2\}$. So $P(E_3=E_1\mid S_1=\{1,2\})=\frac36\cdot\frac1{10}=\frac1{20}.$

Hence $P(S_1=\{1,2\}\cap E_3=E_1)=\frac13\cdot\frac1{20}=\frac1{60}.$

3. Case 2: $S_1=\{1,3\}$

Then: $E_2=\{2\}, \qquad F_2=F_1\cup S_1=\{1,3,4\}.$ So $S_2$ must be one of: $\{1,3\},\{1,4\},\{3,4\}$ with equal probability $\frac13$.

Now, $G_2=G_1\cup S_2=\{2,3,4,5\}\cup S_2.$ Again, only element 1 may be newly added. So:

• if $1\in S_2$, then $G_2=\{1,2,3,4,5\}$;
• if $1\notin S_2$, then $G_2=\{2,3,4,5\}$.

We need $E_3=E_2\cup S_3=\{2\}\cup S_3=\{1,2,3\},$ so necessarily, $S_3=\{1,3\}.$ This requires $1\in G_2$, hence $1\in S_2$.

Among the 3 possible $S_2$, two contain 1: $\{1,3\},\{1,4\}.$ So probability that $1\in S_2$ is $\frac23$.

Then $G_2=\{1,2,3,4,5\}$ and the probability that $S_3=\{1,3\}$ is again $\frac1{\binom52}=\frac1{10}.$ Thus $P(E_3=E_1\mid S_1=\{1,3\})=\frac23\cdot\frac1{10}=\frac1{15}.$

Hence $P(S_1=\{1,3\}\cap E_3=E_1)=\frac13\cdot\frac1{15}=\frac1{45}.$

4. Case 3: $S_1=\{2,3\}$

Then: $E_2=\{1\}, \qquad F_2=F_1\cup S_1=\{1,2,3,4\}.$ Again $S_2$ is any 2-element subset of $\{1,2,3,4\}$, so 6 equally likely possibilities.

Now we need $E_3=E_2\cup S_3=\{1\}\cup S_3=\{1,2,3\},$ so necessarily, $S_3=\{2,3\}.$

But $\{2,3\}\subseteq G_1=\{2,3,4,5\}$ already, so regardless of $S_2$, $G_2\supseteq \{2,3\}.$ Thus $S_3=\{2,3\}$ is always a possible draw.

Now:

• if $1\in S_2$, then $G_2=\{1,2,3,4,5\}$ and $P(S_3=\{2,3\})=\frac1{\binom52}=\frac1{10};$
• if $1\notin S_2$, then $G_2=\{2,3,4,5\}$ and $P(S_3=\{2,3\})=\frac1{\binom42}=\frac16.$

Among the 6 possible $S_2$ from $\{1,2,3,4\}$:

• 3 contain 1,
• 3 do not contain 1.

Therefore