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Step-by-Step Solution
Step 1: Identify the Original Sample Space
The set from which two integers are selected is {1, 2, ..., 11}. Hence, there are 11 integers. The total number of ways to choose 2 integers from these 11 is
$ \binom{11}{2} $. This is the total number of possible pairs (without any condition).
Step 2: Understand the Condition
We are given the condition that the sum of the selected numbers is even. For two numbers to sum to an even number, they must both be even or both be odd. We will focus only on pairs whose sum is even—this forms the reduced sample space.
Step 3: Determine the Reduced Sample Space Size
Count how many ways we can form pairs with both numbers even, and how many ways we can form pairs with both numbers odd.
Even numbers in {1, 2, ..., 11}: 2, 4, 6, 8, 10
There are 5 even numbers total.
Odd numbers in {1, 2, ..., 11}: 1, 3, 5, 7, 9, 11
There are 6 odd numbers total.
Number of ways to choose 2 even numbers = $ \binom{5}{2} $.
Number of ways to choose 2 odd numbers = $ \binom{6}{2} $.
Therefore, the number of favorable pairs (sum even) is
$ \binom{5}{2} + \binom{6}{2} $.
Step 4: Calculate the Number of Even-Even Pairs
The event of interest is that both selected numbers are even. This can be done in $ \binom{5}{2} $ ways.
Step 5: Apply Conditional Probability
The probability we want is:
$ P(\text{both even} \mid \text{sum is even})
= \frac{\binom{5}{2}}{\binom{5}{2} + \binom{6}{2}}. $
Step 6: Simplify and Obtain the Final Answer
$ \binom{5}{2} = \frac{5 \times 4}{2} = 10, \quad \binom{6}{2} = \frac{6 \times 5}{2} = 15. $
So the probability is
$ \frac{10}{10 + 15} = \frac{10}{25} = \frac{2}{5}. $
Conclusion
The required conditional probability that both selected numbers are even, given their sum is even, is
$ \frac{2}{5}. $
