<Question>
Two fair, 6-sided dice are thrown.  What is the probability that the product of the two numbers is a multiple of 5?  Express your answer as a common fraction.
</Question>

<think>
<Plan_1>
Okay, so I need to find the probability that the product of two fair 6-sided dice is a multiple of 5. Hmm, let me think. First, I remember that probability problems often involve figuring out the number of favorable outcomes divided by the total number of possible outcomes. So, maybe I should start by calculating the total number of possible outcomes when two dice are thrown.
Since each die has 6 faces, the total number of outcomes should be 6 multiplied by 6, which is 36. That part seems straightforward.

Now, the tricky part is figuring out how many of these 36 outcomes result in a product that's a multiple of 5. A product being a multiple of 5 means that at least one of the numbers multiplied must be a 5. Because 5 is a prime number, right? So, if either one of the dice shows a 5 or both show a 5, then the product will have 5 as a factor, making it a multiple of 5. Wait, is that correct?

Let me verify.

<prompt_1.1> Verify the logic that if either or both dice show a 5, the product will be a multiple of 5. </prompt_1.1>

So, how do I calculate the number of favorable outcomes here? Maybe I can use the principle of inclusion and exclusion. Let's see.

<prompt_1.2> Use the principle of inclusion and exclusion to calculate the number of favorable outcomes where the product is a multiple of 5. </prompt_1.2>
-----

Alternatively, another way to approach this problem is to calculate the probability that the product is not a multiple of 5, and then subtract that from 1. Sometimes this method is easier.

<prompt_1.3> Calculate the probability that the product is not a multiple of 5 and subtract from 1 to find the desired probability. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
If neither die shows a 5, then the product can't have 5 as a factor. For example, if one die is 1 and the other is 2, the product is 2, which isn't a multiple of 5. But if one die is 5 and the other is anything, say 3, then 5*3=15, which is a multiple of 5. Similarly, 5*5=25, which is also a multiple of 5. So, yes, it seems like the product is a multiple of 5 if and only if at least one of the numbers is 5.
</execution_1.1>

<execution_1.2>
First, the number of outcomes where the first die is 5. If the first die is 5, the second die can be any of the 6 numbers. So that's 6 outcomes. Similarly, the number of outcomes where the second die is 5. The first die can be any of the 6 numbers, and the second die is 5. That's another 6 outcomes. But wait, if I just add those together, I might be double-counting the case where both dice are 5. So, I need to subtract that one overlapping case. So, total favorable outcomes would be 6 + 6 - 1 = 11. Therefore, the probability is 11/36.
</execution_1.2>

<execution_1.3>
The complement of the event "product is a multiple of 5" is "product is not a multiple of 5". For the product not to be a multiple of 5, neither of the numbers can be 5. Because if neither is 5, then the product won't have 5 as a factor. So, if both dice must not show a 5, then each die has 5 possible outcomes (1-4,6). So, the number of such outcomes is 5*5=25. Therefore, the number of outcomes where the product is a multiple of 5 is 36-25=11. Wait, that matches the previous result. So, 11/36. Hmm, that's a good check.

So, either way, whether I calculate directly using inclusion-exclusion or use the complementary probability, I get 11/36. That gives me more confidence that the answer is correct.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2:
But let me check this by enumerating the possibilities to make sure I haven't made a mistake. Let's list all the outcomes where at least one die is 5.

<prompt_2.1> Enumerate the outcomes where at least one die is 5 to verify the calculation of favorable outcomes. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6). That's 6 outcomes. Now, the second die is 5 but the first die isn't 5: (1,5), (2,5), (3,5), (4,5), (6,5). That's another 5 outcomes. Wait, so 6 + 5 = 11. Oh, right! Because when I counted the first die as 5, the second die being 5 was already included in that count. So, to avoid double-counting the (5,5) case, I should subtract 1. So, 6 + 5 - 1 = 10? Wait, no. Wait, when the first die is 5, the second die can be 1-6, including 5. Then, when considering the second die as 5, the first die can be 1-6, but we've already counted when the first die is 5. So, actually, the unique outcomes where at least one die is 5 are:

All pairs where the first die is 5 and the second die is anything: 6. Then, all pairs where the second die is 5 and the first die is not 5: 5 (since (5,5) is already counted). So total 6 + 5 = 11. Therefore, 11 outcomes.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me list them to confirm:

<prompt_3.1> List the outcomes where at least one die is 5 and verify the total number of favorable outcomes. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
First die 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) – 6 outcomes.

Second die 5 but first die not 5: (1,5), (2,5), (3,5), (4,5), (6,5) – 5 outcomes.

Total 6 + 5 = 11. Yes, that's 11 outcomes. So, 11 favorable outcomes out of 36. So, the probability is 11/36.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1, execution_1.3:
Therefore, the probability that the product of the two numbers is a multiple of 5 is 11/36.
I think that's the correct answer.

<prompt_4.1> Conclude the solution by confirming the probability that the product of the two numbers is a multiple of 5. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Both methods of calculation (direct counting with inclusion-exclusion and complementary counting) lead to the same result, so I feel confident about this. I don't see any mistakes in the reasoning here. So, the final answer should be 11/36.
</execution_4.1>
</Execution_4>

<Final_answer>
Therefore, the probability that the product of the two numbers is a multiple of 5 is $\boxed{\frac{11}{36}}$.
</Final_answer>
</think>