Inspired by Nihar and Sandeep
When a non-negative integer \(x\) is divided by 5, we get a remainder \(y\) such that the number \(x\) can be represented as \(x=q×5+y\). For an explicit example if \(x=41\), then we have \(x=8×5+1\). And note that the remainder, here, is 1.
If I give you three chances to enter the integer \(y\), what is the probability that you will get the question right on the third attempt?
You will have to come up with a valid option (a possible remainder which we get after division by 5) each time you enter an integer.
The remainder is defined being bounded in the interval \([0, 5)\)