Two integers are selected at random from the set {1, 2,...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

Question

If \( P(A|B) = P(A'|B) \), then which of the following statements is true?

Answer 1

To solve this problem, we need to find the conditional probability that both chosen numbers are even, given that their sum is even. Let's break down the solution step by step:

Identify the set of integers and classify them as odd or even:

The given set is \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}. From this set:

Even numbers: \{2, 4, 6, 8, 10\}.
Odd numbers: \{1, 3, 5, 7, 9, 11\}.

Determine the total ways to pick two numbers from the set:

The number of ways to choose any two numbers from 11 is given by the combination formula C(n, r) = \frac{n!}{r!(n-r)!}. So,

C(11, 2) = \frac{11 \times 10}{2} = 55

Calculate the favorable outcomes to have an even sum:

To have an even sum, both numbers must be either even or odd.
Number of ways to choose two even numbers: C(5, 2) = \frac{5 \times 4}{2} = 10.
Number of ways to choose two odd numbers: C(6, 2) = \frac{6 \times 5}{2} = 15.

Total number of ways the sum can be even: 10 + 15 = 25

Find the probability that both numbers are even, given that their sum is even:

The conditional probability P(A \mid B) is given by:

P(\text{both numbers even} \mid \text{sum is even}) = \frac{\text{Number of ways to choose two even numbers}}{\text{Total number of ways to have an even sum}}

= \frac{10}{25} = \frac{2}{5}

Conclusion:

The conditional probability that both numbers are even, given that their sum is even, is \frac{2}{5}. Hence, the correct answer is \frac{2}{5}.

Answer 2