To solve the problem, we need to determine the probability that one die shows a multiple of 2 and the other die shows a number that is neither a multiple of 3 nor a multiple of 2.
### Step 1: Identify the outcomes for the first die
The first die must show a multiple of 2. The multiples of 2 on a standard die (which has faces numbered from 1 to 6) are:
- 2
- 4
- 6
Thus, the favorable outcomes for the first die are: **2, 4, 6**.
This gives us a total of **3 favorable outcomes**.
### Step 2: Identify the outcomes for the second die
The second die must show a number that is neither a multiple of 3 nor a multiple of 2. The numbers on a die are 1, 2, 3, 4, 5, and 6.
- Multiples of 2: 2, 4, 6
- Multiples of 3: 3, 6
Now, we need to find the numbers that are neither multiples of 2 nor multiples of 3:
- The numbers on the die are: 1, 2, 3, 4, 5, 6
- Excluding multiples of 2: 1, 3, 5
- Excluding multiples of 3: 1, 2, 3, 4, 5, 6
The only number that remains is **1** and **5**.
Thus, the favorable outcomes for the second die are: **1, 5**.
This gives us a total of **2 favorable outcomes**.
### Step 3: Calculate the total outcomes
When rolling two dice, the total number of outcomes is:
\[
6 \times 6 = 36
\]
(since each die has 6 faces).
### Step 4: Calculate the favorable outcomes
The total favorable outcomes for the event where one die shows a multiple of 2 and the other die shows neither a multiple of 3 nor a multiple of 2 is:
- For the first die (multiple of 2): 3 outcomes (2, 4, 6)
- For the second die (neither multiple of 3 nor 2): 2 outcomes (1, 5)
Thus, the total favorable outcomes = \( 3 \times 2 = 6 \).
### Step 5: Calculate the probability
The probability \( P \) of the event is given by:
\[
P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}
\]
### Final Answer
The probability that one die shows a multiple of 2 and the other die shows neither a multiple of 3 nor a multiple of 2 is \( \frac{1}{6} \).
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