Calculate Conditional Probability of Event in Successive Random Set Selections

To find the conditional probability $p = P(S_1 = \{1, 2\} \mid E_3 = E_1)$, we first analyze the condition under which $E_3 = E_1$.

1. Condition for $E_3 = E_1$

We are given:

• $E_1 = \{1, 2, 3\}$
• $E_2 = E_1 \setminus S_1$
• $E_3 = E_2 \cup S_3$

Since two elements are selected at random from $E_1$ to form $S_1$, $S_1$ can be any 2-element subset of $\{1, 2, 3\}$, and $E_2$ is the remaining 1-element subset.

For $E_3 = E_2 \cup S_3 = \{1, 2, 3\}$ to hold:

1. $S_3$ cannot contain any element outside $\{1, 2, 3\}$ (such as $4$ or $5$), so $S_3 \subseteq \{1, 2, 3\}$.
2. Since $|S_3| = 2$ and $|E_2| = 1$, the union $E_2 \cup S_3$ has size $3$ if and only if $S_3$ contains the two elements of $E_1$ that are not in $E_2$.
3. By definition, $E_1 \setminus E_2 = S_1$. Thus, we must have: $S_3 = S_1$

Hence, $E_3 = E_1$ occurs if and only if $S_3 = S_1$.

---

2. Analysis of Possible Cases for $S_1$

Since 2 elements are chosen at random from $E_1 = \{1, 2, 3\}$, there are $\binom{3}{2} = 3$ equally likely choices for $S_1$, each occurring with probability $\frac{1}{3}$:

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

We compute $P(E_3 = E_1 \text{ and } S_1) = P(S_3 = S_1 \text{ and } S_1)$ for each case.

---

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

• $P(S_1 = \{1, 2\}) = \frac{1}{3}$
• $E_2 = \{3\}$
• $F_2 = F_1 \cup S_1 = \{1, 3, 4\} \cup \{1, 2\} = \{1, 2, 3, 4\}$ (size 4)

Now, 2 elements ($S_2$) are chosen from $F_2$. There are $\binom{4}{2} = 6$ possible choices for $S_2$.

For $S_3 = \{1, 2\}$ to be chosen from $G_2 = G_1 \cup S_2 = \{2, 3, 4, 5\} \cup S_2$, we must have $1 \in G_2$, which requires $1 \in S_2$. The choices for $S_2$ containing $1$ are: $\{1, 2\}$, $\{1, 3\}$, and $\{1, 4\}$ ($3$ choices).

For each of these $3$ choices:

• $G_2 = \{1, 2, 3, 4, 5\}$ (size 5)
• $P(S_3 = \{1, 2\} \mid G_2) = \frac{1}{\binom{5}{2}} = \frac{1}{10}$

Thus, the probability for Case 1 is: $P(E_3 = E_1 \text{ and } S_1 = \{1, 2\}) = \frac{1}{3} \times \left( 3 \times \frac{1}{6} \times \frac{1}{10} \right) = \frac{1}{3} \times \frac{3}{60} = \frac{1}{60}$

---

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

• $P(S_1 = \{1, 3\}) = \frac{1}{3}$
• $E_2 = \{2\}$
• $F_2 = F_1 \cup S_1 = \{1, 3, 4\} \cup \{1, 3\} = \{1, 3, 4\}$ (size 3)

Now, 2 elements ($S_2$) are chosen from $F_2$. There are $\binom{3}{2} = 3$ possible choices for $S_2$.

For $S_3 = \{1, 3\}$ to be chosen from $G_2$, we need $1 \in G_2 \implies 1 \in S_2$. The choices for $S_2$ containing $1$ are: $\{1, 3\}$ and $\{1, 4\}$ ($2$ choices).

For each of these $2$ choices:

• $G_2 = \{1, 2, 3, 4, 5\}$ (size 5)
• $P(S_3 = \{1, 3\} \mid G_2) = \frac{1}{\binom{5}{2}} = \frac{1}{10}$

Thus, the probability for Case 2 is: $P(E_3 = E_1 \text{ and } S_1 = \{1, 3\}) = \frac{1}{3} \times \left( 2 \times \frac{1}{3} \times \frac{1}{10} \right) = \frac{2}{90} = \frac{1}{45}$

---

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

• $P(S_1 = \{2, 3\}) = \frac{1}{3}$
• $E_2 = \{1\}$
• $F_2 = F_1 \cup S_1 = \{1, 3, 4\} \cup \{2, 3\} = \{1, 2, 3, 4\}$ (size 4)

Now, 2 elements ($S_2$) are chosen from $F_2$. There are $\binom{4}{2} = 6$ possible choices for $S_2$.

Since $S_1 = \{2, 3\} \subseteq G_1 = \{2, 3, 4, 5\}$, $S_1$ is already a subset of $G_2$ regardless of $S_2$.

The 6 choices for $S_2$ are divided into two sub-cases:

1. $S_2$ contains $1$ ($\{1, 2\}, \{1, 3\}, \{1, 4\}$ - 3 choices):

   • $G_2 = \{1, 2, 3, 4, 5\}$ (size 5)
   • $P(S_3 = \{2, 3\} \mid G_2) = \frac{1}{\binom{5}{2}} = \frac{1}{10}$
   • Contribution = $3 \times \frac{1}{6} \times \frac{1}{10} = \frac{1}{20}$

2. $S_2$ does not contain $1$ ($\{2, 3\}, \{2, 4\}, \{3, 4\}$ - 3 choices):

   • $G_2 = \{2, 3, 4, 5\}$ (size 4)
   • $P(S_3 = \{2, 3\} \mid G_2) = \frac{1}{\binom{4}{2}} = \frac{1}{6}$
   • Contribution = $3 \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{12}$

Summing these contributions gives: $P(S_3 = \{2, 3\} \mid S_1 = \{2, 3\}) = \frac{1}{20} + \frac{1}{12} = \frac{8}{60} = \frac{2}{15}$

Thus, the probability for Case 3 is: $P(E_3 = E_1 \text{ and } S_1 = \{2, 3\}) = \frac{1}{3} \times \frac{2}{15} = \frac{2}{45}$

---

3. Calculation of Conditional Probability

The total probability of $E_3 = E_1$ is: $P(E_3 = E_1) = P(\text{Case 1}) + P(\text{Case 2}) + P(\text{Case 3})$ $P(E_3 = E_1) = \frac{1}{60} + \frac{1}{45} + \frac{2}{45} = \frac{1}{60} + \frac{3}{45} = \frac{1}{60} + \frac{1}{15} = \frac{5}{60} = \frac{1}{12}$

Using Bayes' theorem, the required conditional probability $p$ is: $p = P(S_1 = \{1, 2\} \mid E_3 = E_1) = \frac{P(E_3 = E_1 \text{ and } S_1 = \{1, 2\})}{P(E_3 = E_1)}$ $p = \frac{1/60}{1/12} = \frac{12}{60} = \frac{1}{5}$

Thus, the value of $p$ is $\frac{1}{5}$, corresponding to option (A).