Number X is randomly selected from the set of odd numbers and Y is randomly selected from the set of even numbers of the set (1.2.3.4.5.6.7). Let. Z = (X + Y).
What is P(Z = 10) equal to?

Options

A. 0
B. `(1)/(2)`
C. `(1)/(3)`
D. `(1)/(5)`

Answer

To solve the problem, we need to find the probability \( P(Z = 10) \) where \( Z = X + Y \), with \( X \) being an odd number selected from the set of odd numbers and \( Y \) being an even number selected from the set of even numbers from the set \( \{1, 2, 3, 4, 5, 6, 7\} \).

### Step-by-Step Solution:

1. **Identify the Sets**:
- The odd numbers from the set \( \{1, 2, 3, 4, 5, 6, 7\} \) are \( \{1, 3, 5, 7\} \).
- The even numbers from the same set are \( \{2, 4, 6\} \).

2. **Determine Possible Values of \( Z \)**:
- Since \( Z = X + Y \), we need to find the combinations of \( X \) (odd) and \( Y \) (even) that can yield a sum of \( Z \).
- The possible sums \( Z \) can take are:
- If \( X = 1 \): \( Z = 1 + 2 = 3 \), \( Z = 1 + 4 = 5 \), \( Z = 1 + 6 = 7 \)
- If \( X = 3 \): \( Z = 3 + 2 = 5 \), \( Z = 3 + 4 = 7 \), \( Z = 3 + 6 = 9 \)
- If \( X = 5 \): \( Z = 5 + 2 = 7 \), \( Z = 5 + 4 = 9 \), \( Z = 5 + 6 = 11 \)
- If \( X = 7 \): \( Z = 7 + 2 = 9 \), \( Z = 7 + 4 = 11 \), \( Z = 7 + 6 = 13 \)

3. **List All Possible Values of \( Z \)**:
- The possible values of \( Z \) from the combinations above are \( \{3, 5, 7, 9, 11, 13\} \).

4. **Check if \( Z = 10 \) is Possible**:
- From the list of possible values of \( Z \), we see that \( 10 \) is not included. This means it is impossible to get \( Z = 10 \) from the sums of an odd number and an even number.

5. **Calculate the Probability**:
- Since \( Z = 10 \) is not a possible outcome, the probability \( P(Z = 10) \) is \( 0 \).

### Final Answer:
Thus, \( P(Z = 10) = 0 \).