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 gt 11)` equal to ?

Options

A. 0
B. `(1)/(4)`
C. `(1)/(6)`
D. `(1)/(12)`

Answer

To solve the problem, we need to find the probability that the sum \( Z = X + Y \) is greater than 11, where \( X \) is an odd number selected from the set of odd numbers in \( \{1, 2, 3, 4, 5, 6, 7\} \) and \( Y \) is an even number selected from the set of even numbers in the same set.

### Step-by-Step Solution:

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

2. **Count the Elements**:
- The total number of odd numbers (for \( X \)) is 4.
- The total number of even numbers (for \( Y \)) is 3.

3. **Calculate Total Possible Outcomes**:
- The total number of possible pairs \( (X, Y) \) is \( 4 \times 3 = 12 \).

4. **Determine the Condition \( Z > 11 \)**:
- We need to find pairs \( (X, Y) \) such that \( X + Y > 11 \).
- The maximum value of \( Z \) occurs when \( X = 7 \) and \( Y = 6 \), giving \( Z = 13 \).
- The only pair that satisfies \( Z > 11 \) is \( (7, 6) \):
- \( 7 + 6 = 13 \) (which is greater than 11).

5. **Count the Favorable Outcomes**:
- There is only 1 favorable outcome: the pair \( (7, 6) \).

6. **Calculate the Probability**:
- The probability \( P(Z > 11) \) is given by the ratio of favorable outcomes to total outcomes:
\[ P(Z > 11) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{12} \]

### Final Answer:
\[ P(Z > 11) = \frac{1}{12} \]