Question

Two integers among 1 to 11 are selected at random. If their sum is even, then find the probability that both integers are odd.

Hint:

For problems involving selecting integers randomly, always count the total possible outcomes and the favorable outcomes to calculate probability.

Answer 1

Step 1: Identify possible combinations.
The sum of two integers is even if both integers are either even or both are odd. In this case, we need to find the probability that both integers selected are odd.
Step 2: Total possible outcomes.
The total number of ways to select 2 integers from the set \( \{1, 2, 3, \dots, 11\} \) is: \[ \binom{11}{2} = \frac{11 \times 10}{2} = 55 \]
Step 3: Favorable outcomes.
The number of odd integers in the set \( \{1, 2, 3, \dots, 11\} \) is 6 (i.e., \( \{1, 3, 5, 7, 9, 11\} \)). The number of ways to select 2 odd integers from these 6 is: \[ \binom{6}{2} = \frac{6 \times 5}{2} = 15 \]
Step 4: Calculate the probability.
The probability that both selected integers are odd is: \[ P(\text{both odd}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{15}{55} = \frac{3}{11} \approx 0.273 \] Thus, the probability is \( \frac{3}{11} \) or approximately 0.273.