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?

**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 \([0, 5)\)

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