Dice Probability Problem

How would you go about finding the probability of rolling 2 one' if you were to roll a die 5 times? This is not related to homework but more of a real life question. I can't exactly figure out how you would solve a problem like this because the pair of one's could appear anywhere within the 5 rolls and not just at the beginning.

Please help. It would be greatly appreciated if you could also explain a bit how you did it.

First you need to figure out how many different combinations are possible with two ones.

1 1 x x x
1 x 1 x x
1 x x 1 x
1 x x x 1
x 1 1 x x
x 1 x 1 x
x 1 x x 1
x x 1 1 x
x x 1 x 1
x x x 1 1

There are 10 combinations possible, each with probability (1/6)^2 (5/6)^3 = 125/6^5
This assumes exactly 2 ones, so each one has 1/6 chance therefore (1/6)^2 and each of the other dice have 5/6 chance of not being a one, therefore (5/6)^3.

Multiply by 10 since there are 10 combinations, so probability is 1250/6^5 = 1250/7776

Jim H made a mathematical error.
5^3 = 125, not 1250.
The correct answer is 125/7776.

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The order won't matter. They are independent events.
You need two rolls that are a 1, and three rolls that are not a 1.
(1/6)*(1/6)*(5/6)*(5/6)*(5/6) = 125/7776 or 1.61%

I agree with Jim H