In throwing of two dice, the number of exhaustive events that '5' will never appear on any one of the dice is?

Options

A. 5
B. 18
C. 25
D. 36

Answer

To solve the problem of finding the number of exhaustive events where '5' will never appear on any one of the two dice when they are thrown, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Total Outcomes**:
When throwing two dice, each die has 6 faces. Therefore, the total number of outcomes when throwing two dice is calculated as:
\[ \text{Total Outcomes} = 6 \times 6 = 36 \]

2. **Identify Outcomes with '5'**:
Next, we need to find the outcomes where at least one die shows a '5'. We can do this by counting the outcomes that include '5' on either die.

- **Outcomes with '5' on the First Die**:
- If the first die shows '5', the second die can show any of the 6 numbers (1, 2, 3, 4, 5, 6). This gives us 6 outcomes:
\[ (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \]

- **Outcomes with '5' on the Second Die**:
- If the second die shows '5', the first die can also show any of the 6 numbers. This gives us another 6 outcomes:
\[ (1,5), (2,5), (3,5), (4,5), (5,5), (6,5) \]

- **Overlap**:
- The outcome (5,5) is counted in both cases, so we need to subtract this overlap.

- **Total Outcomes with '5'**:
\[ \text{Total Outcomes with '5'} = 6 + 6 - 1 = 11 \]

3. **Calculate Outcomes Where '5' Never Appears**:
To find the number of outcomes where '5' does not appear on either die, we subtract the number of outcomes that include '5' from the total outcomes:
\[ \text{Outcomes without '5'} = \text{Total Outcomes} - \text{Outcomes with '5'} \]
\[ \text{Outcomes without '5'} = 36 - 11 = 25 \]

### Final Answer:
Thus, the number of exhaustive events where '5' will never appear on any one of the dice is **25**.