Consider three sets $$E_1 = \left\{1, 2, 3 \right\}, F_1 = \left\{1, 3, 4 \right\}$$ and $$G_1 = \left\{2, 3, 4, 5 \right\}$$. Two elements are chosen at random, without replacement, from the set $$E_1$$, and let $$S_1$$ denote the set of these chosen elements. Let $$E_2 = E_1 − S_1$$ and $$F_2 = F_1 \cup S_1$$. Now two elements are chosen at random, without replacement, from the set $$F_2$$ and let $$S_2$$ denote the set of these chosen elements.

Let $$G_2 = G_1 \cup S_2$$. Finally, two elements are chosen at random, without replacement, from the set $$G_2$$ and let $$S_3$$ denote the set of these chosen elements. Let $$E_3 = E_2 \cup S_3$$. Given that $$E_1 = E_3$$, let p be the conditional probability of the event $$S_1 = \left\{1, 2 \right\}$$. Then the value of p is

All selections are made uniformly at random and without replacement. Throughout the solution write $$E_1=\{1,2,3\},\;F_1=\{1,3,4\},\;G_1=\{2,3,4,5\}$$.

Step 1 - The first draw
Two elements are chosen from $$E_1$$. The possible sets are $$\{1,2\},\;\{1,3\},\;\{2,3\}$$ each with probability $$\tfrac13$$. Denote the chosen pair by $$S_1$$. After removing these two elements, $$E_2=E_1-S_1$$ contains exactly the one element of $$E_1$$ that was not chosen.

Step 2 - Updating $$F_2$$ and drawing $$S_2$$
Form $$F_2=F_1\cup S_1$$ and then choose two elements $$S_2$$ from $$F_2$$.

Step 3 - Updating $$G_2$$ and drawing $$S_3$$
Put $$G_2=G_1\cup S_2$$ and choose two elements $$S_3$$ from $$G_2$$.
Finally $$E_3=E_2\cup S_3$$.

Key observation
Because $$E_2$$ already contains the single element of $$E_1$$ which is not in $$S_1$$, the equality $$E_3=E_1$$ can hold only if $$S_3=S_1$$. Thus $$E_1=E_3 \;\Longleftrightarrow\; S_3=S_1.$$

We must calculate $$P\!\left(S_1=\{1,2\}\,\big|\,S_3=S_1\right) =\dfrac{P\!\left(S_1=\{1,2\}\;\text{and}\;S_3=S_1\right)} {P\!\left(S_3=S_1\right)}.$$

Case 1: $$S_1=\{1,2\}$$
$$F_2=\{1,2,3,4\}$$ (4 elements). There are $$\binom42=6$$ possible $$S_2$$; exactly three of them (those containing 1) put 1 into $$G_2$$. Hence $$P(1\in S_2)=\tfrac12.$$ If 1 is in $$S_2$$ then $$G_2=\{1,2,3,4,5\}$$ (5 elements) and $$P(S_3=\{1,2\}\mid1\in S_2)=\tfrac1{10}.$$ Therefore $$P(S_3=S_1\mid S_1=\{1,2\})=\tfrac12\cdot\tfrac1{10}=\tfrac1{20}.$$ Multiplying by $$P(S_1=\{1,2\})=\tfrac13$$ gives $$P(S_1=\{1,2\}\;\text{and}\;S_3=S_1)=\tfrac1{60}.$$

Case 2: $$S_1=\{1,3\}$$
$$F_2=\{1,3,4\}$$ (3 elements). There are $$\binom32=3$$ choices for $$S_2$$; two of them contain 1. Thus $$P(1\in S_2)=\tfrac23$$ and again $$G_2=\{1,2,3,4,5\}$$ giving $$P(S_3=\{1,3\}\mid1\in S_2)=\tfrac1{10}.$$ Hence $$P(S_3=S_1\mid S_1=\{1,3\})=\tfrac23\cdot\tfrac1{10}=\tfrac1{15},$$ so $$P(S_1=\{1,3\}\;\text{and}\;S_3=S_1)=\tfrac13\cdot\tfrac1{15}=\tfrac1{45}.$$

Case 3: $$S_1=\{2,3\}$$
$$F_2=\{1,2,3,4\}$$ (4 elements, 6 pairs). Half of the pairs contain 1, so $$P(1\in S_2)=\tfrac12$$.
• If $$1\in S_2$$ then $$G_2=\{1,2,3,4,5\}$$ and $$P(S_3=\{2,3\})=\tfrac1{10}.$$ • If $$1\notin S_2$$ then $$G_2=\{2,3,4,5\}$$ and $$P(S_3=\{2,3\})=\tfrac1{6}.$$ Thus $$P(S_3=S_1\mid S_1=\{2,3\})= \tfrac12\cdot\tfrac1{10}+\tfrac12\cdot\tfrac1{6} =\tfrac1{20}+\tfrac1{12}=\tfrac2{15},$$ and $$P(S_1=\{2,3\}\;\text{and}\;S_3=S_1)=\tfrac13\cdot\tfrac2{15}=\tfrac2{45}.$$

Total probability of $$S_3=S_1$$
$$P(S_3=S_1)=\tfrac1{60}+\tfrac1{45}+\tfrac2{45} =\tfrac3{180}+\tfrac4{180}+\tfrac8{180} =\tfrac1{12}.$$

Conditional probability required
$$P\!\left(S_1=\{1,2\}\,\big|\,S_3=S_1\right) =\dfrac{\tfrac1{60}}{\tfrac1{12}} =\dfrac{12}{60} =\tfrac15.$$

Hence the desired probability is $$\boxed{\tfrac15}$$.
Option A is correct.