OK. We assume that the last three digits on the bills can be treated as random and "uniformly distributed". This just means that on any bill you pick out, the 1000 possible combinations of 3 digits (running from up to ).
For part A, imagine looking at one bill first - it gives you a specific 3 digits. Then look at the other bill - the probability of those 3 digits being equal to those on the first bill is just:
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.
So, the overall probability of no matches comes by multiplying these nine probabilities together, giving you
- 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 . 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 (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 .
So, the probability of that there is a match (2 or more codes being equal) is .
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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