Question:

Step 1: Classify the numbers into odd and even categories. The given set of numbers is $S = \{1, 2, 3, 5, 7, 11, 12, 13, 17, 19\}$. We first categorize these numbers into odd and even sets: Odd numbers: $O = \{1, 3, 5, 7, 11, 13, 17, 19\}$. The count of odd numbers is $n(O) = 8$. Even numbers: $E = \{2, 12\}$. The count of even numbers is $n(E) = 2$. The total number of elements in the set is $n(S) = 10$. Step 2: Determine the total number of pairs with an even sum. For the sum of two numbers to be even, both numbers must be odd, or both numbers must be even. The pairs are picked without replacement. The number of ways to choose 2 odd numbers from the 8 odd numbers is: $$ \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 $$ The number of ways to choose 2 even numbers from the 2 even numbers is: $$ \binom{2}{2} = 1 $$ The total number of pairs whose sum is even is the sum of these possibilities: $$ \text{Total pairs with even sum} = 28 + 1 = 29 $$ Let $A$ be the event that the sum of the picked numbers is even. So, $n(A) = 29$. Step 3: Determine the number of pairs that include 11 and have an even sum. We need to find the number of pairs where 11 was picked AND the sum is even. Since 11 is an odd number, for the sum of the two picked numbers to be even, the other number picked must also be odd. The original set of odd numbers is $O = \{1, 3, 5, 7, 11, 13, 17, 19\}$. If 11 is already picked, we need to choose one more odd number from the remaining odd numbers. The remaining odd numbers are $O' = \{1, 3, 5, 7, 13, 17, 19\}$. The number of remaining odd numbers is $n(O') = 7$. The number of pairs that include 11 and have an even sum is the number of ways to pick one number from these 7 remaining odd numbers: $$ \binom{7}{1} = 7 $$ Let $B$ be the event that the number 11 was picked. The number of outcomes in the intersection of events $A$ and $B$ (i.e., 11 was picked AND the sum is even) is $n(A \cap B) = 7$. Step 4: Calculate the conditional probability. We need to find the probability that 11 was picked given that the sum of the numbers was even. This is denoted as $P(B|A)$. Using the formula for conditional probability: $$ P(B|A) = \frac{n(A \cap B)}{n(A)} $$ From Step 2, $n(A) = 29$ (total pairs with even sum). From Step 3, $n(A \cap B) = 7$ (pairs including 11 with an even sum). Therefore, the probability is: $$ P(\text{11 picked | sum is even}) = \frac{7}{29} $$ Calculating the numerical value: $$ \frac{7}{29} \approx 0.241379... $$ Rounding to two decimal places, the probability is approximately $0.24$. The final answer is $\boxed{\text{0.24}}$.