Okay, so I need to find the probability that when two fair 6-sided dice are thrown, the product of the two numbers is a multiple of 5. Hmm, let me think about this step by step. 

First, probability problems often involve figuring out the number of favorable outcomes divided by the total number of possible outcomes. Since we're dealing with two dice, each die has 6 faces, so the total number of possible outcomes when throwing two dice is 6 * 6 = 36. That part seems straightforward.

Now, the key here is to determine how many of these 36 outcomes result in a product that's a multiple of 5. A product is a multiple of 5 if at least one of the numbers rolled is a 5. Wait, is that right? Let me check. If either die shows a 5, then the product will definitely be a multiple of 5. But what if neither die shows a 5? Then the product can't be a multiple of 5. So yes, the only way the product isn't a multiple of 5 is if neither die is a 5. 

Therefore, to find the number of favorable outcomes, maybe it's easier to calculate the total outcomes minus the outcomes where neither die is a 5. Let me verify that approach. 

Total outcomes: 36. Outcomes where neither die is a 5: each die has 5 possible outcomes (1-4, 6), so 5 * 5 = 25. Therefore, the number of favorable outcomes is 36 - 25 = 11. So the probability would be 11/36. 

Wait, but let me double-check this because sometimes when you subtract these, you might be missing something. Let's list out the possible pairs where at least one die is 5. 

If the first die is 5, the possible pairs are (5,1), (5,2), (5,3), (5,4), (5,5), (5,6). That's 6 outcomes. If the second die is 5, the possible pairs are (1,5), (2,5), (3,5), (4,5), (5,5), (6,5). That's another 6 outcomes. But wait, the pair (5,5) is counted twice here. So total favorable outcomes are 6 + 6 - 1 = 11. 

Yes, that matches the previous result. So 11 favorable outcomes out of 36. Therefore, the probability is 11/36. 

But let me think again. Is there another way to approach this? Maybe by considering the probability of each die being a 5 and then using inclusion-exclusion principle. 

The probability that the first die is 5 is 1/6. The probability that the second die is 5 is also 1/6. The probability that both are 5 is 1/6 * 1/6 = 1/36. So using inclusion-exclusion, the probability that at least one die is 5 is P(A) + P(B) - P(A and B) = 1/6 + 1/6 - 1/36 = 2/6 - 1/36 = 12/36 - 1/36 = 11/36. 

Same result. So that reinforces that the answer is 11/36. 

Alternatively, maybe I can list all the possible pairs where the product is a multiple of 5. Let's see. 

Numbers on a die are 1, 2, 3, 4, 5, 6. For the product to be a multiple of 5, at least one of the numbers must be 5. So the pairs where either the first or the second die is 5. 

If the first die is 5, the second can be 1,2,3,4,5,6. That's 6 pairs. If the second die is 5, the first can be 1,2,3,4,6 (since (5,5) is already counted). That's 5 additional pairs. So total 6 + 5 = 11. Yep, same as before. 

So all methods lead to 11 favorable outcomes. Therefore, the probability is 11/36. 

I think that's solid. Let me check if there's any other angle. For instance, suppose someone argues that maybe both numbers need to be 5, but no, that's not the case. If either is 5, the product is a multiple of 5. So even if one is 5 and the other is something else, it still works. 

Another way to confirm: list all the pairs where the product is a multiple of 5. 

First die is 5:
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6

Second die is 5, excluding (5,5) which is already counted:
(1,5), (2,5), (3,5), (4,5), (6,5) → 5

Total 11. 

Yes, so that's correct. Therefore, the answer is 11/36. I can't see any mistakes in this reasoning. All approaches converge to the same answer. So I'm confident that's the correct probability.
Therefore, the probability that the product of the two numbers is a multiple of 5 is $\boxed{\frac{11}{36}}$.
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