A fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? Express your answer as a common fraction.

I think the last 3 digits have to be divisible by 8 to be a multiple of 8...

You are correct! To determine if the eight-digit number is divisible by 8, we need to check if the last three digits form a multiple of 8.

To find the probability that the last three digits form a multiple of 8, we need to calculate the total number of favorable outcomes and the total number of possible outcomes.

The total number of possible outcomes is given by the number of ways the six-sided die can be rolled eight times, which is 6^8 since each roll has six possible outcomes.

Next, let's consider the favorable outcomes. For a number to be divisible by 8, the last three digits must form a multiple of 8. Since each digit can range from 1 to 6 (inclusive), there are 6^3 = 216 possible combinations for the last three digits.

However, not all of these combinations will form a multiple of 8. To form a multiple of 8, the last three digits must be one of the following: 016, 024, 032, 064, 072, 104, 112, 136, 144, 152, 176, 184, 208, 216, 232, 256, 264, 272, 304, 312, 336, 344, 352, 376, 384, 416, 424, 456, 464, 472, 504, 512, 536, 544, 552, 576, 584, 616, 624, 632, 664, 672.

There are a total of 43 favorable outcomes.

Thus, the probability of obtaining a multiple of 8 as an eight-digit number is given by:

Probability = Favorable outcomes / Total outcomes

Probability = 43 / (6^8)

Simplifying further:

Probability = 43 / 1679616

Therefore, the probability that the resulting number is a multiple of 8 is 43/1679616.