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 qu​estion right on the third attempt?

Clarification:

• 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 $[0, 5)$