Key Concepts and Formulas

• Conditional Probability: The probability of an event B occurring, given that event A has already occurred, is denoted as $P(B|A)$. It is defined by the formula: $P(B|A) = \frac{P(A \cap B)}{P(A)}$ When all outcomes are equally likely, this can be expressed in terms of the number of favorable outcomes: $P(B|A) = \frac{\text{Number of outcomes in } (A \cap B)}{\text{Number of outcomes in } A} = \frac{N(A \cap B)}{N(A)}$ This means our sample space is effectively reduced to only those outcomes where event A occurs.
• Combinations: The number of ways to choose $r$ items from a set of $n$ distinct items, where the order of selection does not matter, is given by the combination formula: ${}^nC_r = \frac{n!}{r!(n-r)!}$
• Parity Rules for Sums:
  • Even + Even = Even
  • Odd + Odd = Even
  • Even + Odd = Odd

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Step-by-Step Solution

1. Analyze the Given Set and Classify Numbers by Parity The problem involves selecting numbers from the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. There are a total of 11 integers in this set. To determine the parity of sums, it's crucial to first identify the count of even and odd numbers within this set.

• Even numbers (E): These are $\{2, 4, 6, 8, 10\}$. There are $N_E = 5$ even numbers.
• Odd numbers (O): These are $\{1, 3, 5, 7, 9, 11\}$. There are $N_O = 6$ odd numbers.

2. Define the Events Let's clearly define the events involved in the conditional probability question:

• Event A: The sum of the two selected numbers is even. This is the condition that is "given".
• Event B: Both the selected numbers are even. This is the event whose probability we need to find, conditional on A.

We are tasked with finding $P(B|A)$.

3. Determine the Conditions for Event A (Sum is Even) For the sum of two integers to be an even number, based on the parity rules, there are only two possible scenarios for the parities of the two selected numbers:

• Case 1: Both numbers selected are even (Even + Even = Even).
• Case 2: Both numbers selected are odd (Odd + Odd = Even).

Any other combination (one even and one odd) would result in an odd sum.

4. Calculate the Number of Outcomes for Event A ($N(A)$) Event A is the condition that the sum of the two selected numbers is even. We need to count the total number of ways to select two numbers such that their sum is even. This will form our reduced sample space.

• Number of ways to select two even numbers: We have 5 even numbers, and we need to choose 2 of them. This is calculated using combinations: ${}^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$
• Number of ways to select two odd numbers: We have 6 odd numbers, and we need to choose 2 of them: ${}^6C_2 = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$

The total number of ways for Event A to occur (i.e., the sum of the selected numbers is even) is the sum of these two cases: $N(A) = \text{ways to choose 2 even} + \text{ways to choose 2 odd}$ $N(A) = {}^5C_2 + {}^6C_2 = 10 + 15 = 25$ So, there are 25 possible pairs of numbers whose sum is even. This is the size of our reduced sample space.

5. Calculate the Number of Outcomes for Event A $\cap$ B ($N(A \cap B)$) Event $A \cap B$ represents the scenario where "the sum of the two selected numbers is even AND both selected numbers are even." If both selected numbers are even, their sum will always be even (as per parity rules: Even + Even = Even). Therefore, the condition "sum is even" is automatically satisfied if "both numbers are even". Thus, the event $A \cap B$ simplifies to just "both selected numbers are even".

The number of ways to select two even numbers was already calculated in Step 4: $N(A \cap B) = {}^5C_2 = 10$ So, there are 10 pairs of numbers where both are even (and consequently, their sum is even).

6. Calculate the Conditional Probability Now we apply the conditional probability formula using the counts of outcomes: $P(B|A) = \frac{N(A \cap B)}{N(A)}$ Substitute the values we calculated: $P(B|A) = \frac{10}{25}$

7. Simplify the Result Simplify the fraction to its simplest form: $P(B|A) = \frac{10 \div 5}{25 \div 5} = \frac{2}{5}$

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Common Mistakes & Tips

• Reduced Sample Space: A common mistake is to use the total number of ways to select two numbers from the set (which is ${}^{11}C_2 = 55$) as the denominator. Remember that for conditional probability $P(B|A)$, the denominator is $N(A)$, the number of outcomes in the given event A.
• Parity Rules: Always confirm the parity rules for sums. A quick check (e.g., $2+4=6$ (E+E=E), $1+3=4$ (O+O=E), $1+2=3$ (O+E=O)) can prevent errors.
• Combinations vs. Permutations: Since the problem asks for "two integers selected," and the order of selection does not change the pair (e.g., {2,4} is the same as {4,2}), combinations ($^nC_r$) are the correct counting method, not permutations ($^nP_r$).

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Summary

This problem is a straightforward application of conditional probability. We began by classifying the numbers in the given set into even and odd counts. Then, we identified the conditions under which the sum of two selected numbers would be even (either both even or both odd) to determine the size of our restricted sample space, $N(A)$. Next, we determined the number of outcomes where both numbers are even (which also satisfies the condition that their sum is even) to find $N(A \cap B)$. Finally, by taking the ratio $N(A \cap B) / N(A)$, we calculated the conditional probability.

The final answer is $\boxed{{2 \over 5}}$, which corresponds to option (A).