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 numbers X and Y being even, given (X+Y)=15?

Options

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

Answer

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 are drawing two numbers \(X\) and \(Y\) from the set \(\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\). We need to find pairs \((X, Y)\) such that \(X + Y = 15\).

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

So, the sample space \(S\) is:
\[ S = \{(5, 10), (6, 9), (7, 8), (8, 7), (9, 6), (10, 5)\} \]

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

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

The favorable outcomes are:
- \( (5, 10) \)
- \( (6, 9) \)
- \( (9, 6) \)
- \( (10, 5) \)

So, the favorable outcomes \(F\) are:
\[ F = \{(5, 10), (6, 9), (9, 6), (10, 5)\} \]

5. **Count the Number of Favorable Outcomes**:
The number of favorable outcomes is \(4\).

6. **Calculate the Conditional Probability**:
The conditional probability \(P\) of exactly one of the two numbers being even, given that \(X + Y = 15\) is given by:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes in sample space}} = \frac{4}{6} = \frac{2}{3} \]

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