If you roll a 6-sided die 4 times, what is the probability of one roll being a multiple of 3?

See the answer under your first post.

There is a 2/3 chance each time that the roll will NOT be a multiple of 3, so (2/3)^4 is the probability that none of them will be a multiple of 3.

If we subtract this from 1, we will get the probablility that at least one of the rolls is a multiple of 3. This is equal to 65/81, option D.

I hope this helps you, best of luck!

This is exactly the reverse of your first question about the Pirates! Out of 6^4 =1,296 permutations, there are 1040 of them whose product is a multiple of 3. Or, 1040 / 6^4 =65 /81