Consider three sets $E_1 = \{1, 2, 3\}$, $F_1 = \{1, 3, 4\}$ and $G_1 = \{2, 3, 4, 5\}$. 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 $S_1 = \{1, 2\}$. Then the value of $p$ is

Consider three sets E₁ = {1, 2, 3}, F₁ = {1, 3, 4} and G₁ = {2, 3, 4, 5}. Two elements are chosen at random, without replacement, from the set E₁, and let S₁ denote the set of these chosen elements. Let E₂ = E₁ - S₁ and F₂ = F₁ ∪ S₁. Now two elements are chosen at random, without replacement, from the set F₂ and let S₂ denote the set of these chosen elements. Let G₂ = G₁ ∪ S₂. Finally, two elements are chosen at random, without replacement, from the set G₂ and let S₃ denote the set o

$E_1=\{1,2,3\},\quad F_1=\{1,3,4\},\quad G_1=\{2,3,4,5\}$

$S_1 \subset E_1,\ |S_1|=2;\quad E_2 = E_1 - S_1,\quad F_2 = F_1 \cup S_1$

$S_2 \subset F_2,\ |S_2|=2;\quad G_2 = G_1 \cup S_2$

$S_3 \subset G_2,\ |S_3|=2;\quad E_3 = E_2 \cup S_3$

$\text{Given } E_1 = E_3:\quad p = P\!\left(S_1 = \{1,2\} \mid E_1 = E_3\right)$

$\text{To find: } p$

$S_1 \text{ is one of } \{1,2\},\ \{1,3\},\ \{2,3\},\ \text{each with prob } \dfrac{1}{3}$

$E_3 = E_2 \cup S_3 = E_1 \iff S_3 \supseteq E_1 - E_2 = S_1$

$\iff S_3 = S_1 \quad (\text{both have 2 elements})$

$\text{Plan: } P(E_1 = E_3) = \sum_{S_1} P(S_1)\,P(S_3 = S_1 \mid S_1)$

$p = \dfrac{P(S_1 = \{1,2\},\, S_3 = S_1)}{P(S_3 = S_1)}$

$S_1=\{1,2\}:\ E_2=\{3\},\ F_2=\{1,2,3,4\},\ \text{need } S_3=\{1,2\}$

$P(1\in S_2)=\dfrac{3}{6}=\dfrac{1}{2};\quad 1\in S_2 \Rightarrow G_2=\{1,2,3,4,5\}$

$P(S_3=S_1)=\dfrac{1}{2}\cdot\dfrac{1}{\binom{5}{2}}=\dfrac{1}{2}\cdot\dfrac{1}{10}=\dfrac{1}{20}$

$S_1=\{1,3\}:\ E_2=\{2\},\ F_2=\{1,3,4\},\ \text{need } S_3=\{1,3\}$

$S_2 \in \{\{1,3\},\{1,4\},\{3,4\}\};\quad P(1 \in S_2)=\dfrac{2}{3}$

$P(S_3=S_1)=\dfrac{2}{3}\cdot\dfrac{1}{10}=\dfrac{1}{15}$

$S_1=\{2,3\}:\ E_2=\{1\},\ F_2=\{1,2,3,4\},\ \text{need } S_3=\{2,3\}$

$2,3 \in G_2 \text{ always};\quad P(1 \in S_2)=\dfrac{1}{2}$

$P(S_3=S_1)=\dfrac{1}{2}\cdot\dfrac{1}{10}+\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{1}{20}+\dfrac{1}{12}=\dfrac{2}{15}$

$P(S_3 = S_1) = \dfrac{1}{3}\left(\dfrac{1}{20}+\dfrac{1}{15}+\dfrac{2}{15}\right)$

$= \dfrac{1}{3}\cdot\dfrac{1}{4} = \dfrac{1}{12}$

$P(S_1 = \{1,2\},\, S_3 = S_1) = \dfrac{1}{3}\cdot\dfrac{1}{20} = \dfrac{1}{60}$

$p = \dfrac{P(S_1 = \{1,2\},\, S_3 = S_1)}{P(S_3 = S_1)} = \dfrac{1/60}{1/12} = \dfrac{12}{60} = \dfrac{1}{5}$

$p = \dfrac{1}{5} \quad \Rightarrow \quad \text{option (1)}$