Part B is a little more complicated. To solve it, I'm going to ask a slightly different question:

*"You have 10 bills. What is the probability that ***no two** of them share their last three digits?"

This time, imagine laying all 10 bills out in a row. Starting at one end, read the bills one at a time. Stop whenever you see a "match". If you get to the end without reading the same number twice, then (obviously) no two of them are the same.

- When you look at "bill #1", you get one 3-digit code to hold in your head.
- When you look at "bill #2", the probability of that code not being equal to bill #1 is $\displaystyle 999/1000$. And, in this case, you can move on to bill #3.
- When you look at "bill #3", the probability of it not being equal to
**either bill #1 or bill #2** is $\displaystyle 998/1000$ (by the time to get to reading bill #3, you know that bills #1 & 2 are different.) - Keep going. If you make it to bill #10, you'll have 9 distinct numbers in your head, and the probability that bill #10 isn't equal to any of them is $\displaystyle 991/1000$.

So, the overall probability of no matches comes by multiplying these nine probabilities together, giving you

$\displaystyle P(no~match) = (1 - P_{10}) = \frac{999}{1000} \times \frac{998}{1000} \times \ldots \times \frac{991}{1000} \approx 0.956$.

So, the probability of that there

**is** a match (2 or more codes being equal) is $\displaystyle P_{10} = 1 - 0.956 = 0.044 = 4.4\%$.

- - -

I fear I might have made this look more complicated than it really is. I invite anyone with a more concise argument to jump in. ;P