Two numbers X and Y are simultaneously drawn from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. What is the conditional probability of exactly one of the two numebrs X and Y being even, given (X + Y) = 15?

Options

A. 1
B. `3//4`
C. `1//2`
D. `1//4`

Explanation

To solve the problem, we need to find the conditional probability of exactly one of the two numbers \(X\) and \(Y\) being even, given that \(X + Y = 15\).

### Step-by-Step Solution:

1. **Identify the Sample Space:**
We start by identifying all pairs \((X, Y)\) such that \(X + Y = 15\) and both \(X\) and \(Y\) are drawn from the set \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\).

2. **List the Valid Pairs:**
The valid pairs that satisfy \(X + Y = 15\) are:
- \( (5, 10) \)
- \( (6, 9) \)
- \( (7, 8) \)
- \( (8, 7) \)
- \( (9, 6) \)
- \( (10, 5) \)

Thus, the sample space \(S\) consists of these 6 pairs.

3. **Count the Total Outcomes:**
The total number of outcomes in the sample space \(S\) is \(6\).

4. **Identify the Favorable Outcomes:**
We need to find pairs where exactly one of the numbers is even.
- From the pairs:
- \( (5, 10) \): One even (10)
- \( (6, 9) \): One even (6)
- \( (7, 8) \): One even (8)
- \( (8, 7) \): One even (8)
- \( (9, 6) \): One even (6)
- \( (10, 5) \): One even (10)

The pairs where exactly one number is even are:
- \( (5, 10) \)
- \( (6, 9) \)
- \( (9, 6) \)
- \( (10, 5) \)

Therefore, there are \(4\) favorable outcomes.

5. **Calculate the Conditional Probability:**
The conditional probability \(P(A|B)\) is given by the formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Here, \(P(A \cap B)\) is the number of favorable outcomes (4) and \(P(B)\) is the total outcomes (6).

Thus, the conditional probability is:
\[ P(\text{exactly one even} | X + Y = 15) = \frac{4}{6} = \frac{2}{3} \]

### Final Answer:
The conditional probability of exactly one of the two numbers \(X\) and \(Y\) being even, given that \(X + Y = 15\), is \(\frac{2}{3}\).

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