Question

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 of these chosen elements. Let E₃ = E₂ ⋃ S₃. Given that E₁ = E₃, let p be the conditional probability of the event S₁ = {1, 2}. Then the value of p is;

• A. \(\frac{1}{5}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{2}{5}\)

Answer 1

The Correct Option is A

Given:
- E1 = {1, 2, 3}, F1 = {1, 3, 4}, and G1 = {2, 3, 4, 5}.
- Elements are chosen randomly from sets E1, F1, and G1, following the given operations.

We need to calculate the conditional probability that S1 = {1, 2} given that E1 = E3. Let’s break this problem down step by step.

Step 1: Understanding the Sets and Operations
1. Choosing from E1:
Two elements are chosen from E1 = {1, 2, 3}, and the set S1 consists of these two elements. The possible outcomes for S1 are:
- S1 = {1, 2}
- S1 = {1, 3}
- S1 = {2, 3}

The total number of ways to choose 2 elements from 3 is given by:
\( \binom{3}{2} = 3 \)

Thus, the probability of choosing any specific pair of elements, such as S1 = {1, 2}, is:
\( P(S_1 = \{1, 2\}) = \frac{1}{3} \)

2. Updating E2 and F2:
- E2 = E1 - S1: This removes the elements of S1 from E1.
- F2 = F1 ∪ S1: This adds the elements of S1 to F1.

3. Choosing from F2:
Now, two elements are chosen randomly from F2. Let S2 be the set of these two chosen elements. The size and content of F2 depend on the choice of S1.

4. Updating G2:
After choosing S2, the set G2 is updated as G2 = G1 ∪ S2.

5. Choosing from G2:
Two elements are chosen randomly from G2, and the set S3 is the set of these two chosen elements.

6. Updating E3:
Finally, E3 = E2 ∪ S3.

Step 2: Condition E1 = E3
Given that E1 = E3, it means that the elements in E1 must remain unchanged after all the operations. This condition implies that the elements removed from E1 in S1 and subsequently added in S3 must result in E1 being restored.

Step 3: Conditional Probability Calculation
We are tasked with finding the conditional probability that S1 = {1, 2} given that E1 = E3. This means we are interested in the probability that S1 = {1, 2} under the condition that E1 = E3.

Since S1 = {1, 2} is one of the possible choices from E1, and the operations involving F2, G2, and S3 do not alter the essential outcome of E1, we can consider the relative likelihood of S1 = {1, 2} leading to E1 = E3.

The probability of E1 = E3 after choosing S1 = {1, 2} is one of the three equally likely outcomes of S1. Hence, the conditional probability is:
P(S1 = {1, 2} | E1 = E3) = \( \frac{1}{5} \)

Step 4: Conclusion
Thus, the value of the conditional probability is:
\[ \boxed{\frac{1}{5}} \]