Problem Analysis: We need to find the conditional probability that two integers selected randomly from the first 11 natural numbers are both odd, given that their sum is even.

Identify Number Properties

The first 11 natural numbers are $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.

• Total numbers = 11
• Odd numbers (O) = $\{1, 3, 5, 7, 9, 11\}$. Count = 6.
• Even numbers (E) = $\{2, 4, 6, 8, 10\}$. Count = 5.

Determine Condition for Even Sum

The sum of two integers is even if:

• Both integers are odd (O + O = Even).
• Both integers are even (E + E = Even).

Calculate Ways for Even Sum

We calculate the number of ways to select pairs that satisfy the condition (sum is even).

• Ways to select 2 odd numbers from 6: $N_{OO} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$
• Ways to select 2 even numbers from 5: $N_{EE} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$
• Total ways to select 2 numbers with an even sum: $N_{\text{Sum Even}} = N_{OO} + N_{EE} = 15 + 10 = 25$

This forms our reduced sample space.

Calculate Favorable Outcomes

We are interested in the event where both selected numbers are odd. The number of ways for this event to occur within the reduced sample space is $N_{OO}$.

• Favorable outcomes (both numbers are odd) = $N_{OO} = 15$.

Compute Conditional Probability

The conditional probability is the ratio of the number of favorable outcomes to the total number of outcomes in the reduced sample space.

Probability (Both Odd | Sum Even) = $ \frac{\text{Number of ways both are odd}}{\text{Total number of ways sum is even}} $

Probability = $ \frac{N_{OO}}{N_{\text{Sum Even}}} = \frac{15}{25} $

Simplifying the fraction:

Probability = $ \frac{15}{25} = \frac{3}{5} $

Final Answer

The probability that both numbers are odd, given their sum is even, is $ \frac{3}{5} $.