Each vehicle has a number plate: 0001-9999, what is the chance that 2 random cars will have identical digit in their plate?

Question

What is the chance that there won't be a identical digit in two random plates if there are totally $9999$ plates for which number sequence varies from $0001 - 9999.$ So, the chance that two random plates will have 0 in them is $(10^4-9^4)^2= 11826721 $ The same goes with other numbers 1-9. Should I simply multiply $(10^4-9^4) \cdot 10$?

Answer 1

The easiest approach in my opinion is to first make the simplification that we allow the license plate to be $0000$ as well., making it a clean $10^4$ possibilities where we don't need to give zero any special treatment until the very end where we subtract this away as having been a possibility.

We continue counting the number of ways of having two plates sharing at least one similar digit between them with inclusion-exclusion based on the events "Both plates have at least one zero", "Both plates have at least one 1", "Both plates have at least one 2", etc...

The number of plates containing at least one zero is equal to the number of plates minus the number of plates which contain no zeroes and would be $10^4-9^4$. The number of ways that both plates have at least one zero would then be $(10^4-9^4)^2$

The number of plates containing at least one zero and at least one $1$ can be counted again by breaking down with inclusion exclusion as the number of plates minus those plates which are missing at least one of 0 or 1 and would be $10^4-9^4-9^4+8^4$ making the number of ways that both plates simultaneously contain at least one $0$ and at least one $1$ as being $(10^4-9^4-9^4+8^4)^2$

Continue in this fashion and correctly apply inclusion-exclusion across all of these events to get the count. Then, remove from your count the possibilities where either user had the "bad" license plate of $0000$ and the other had a $0$.

Finally, divide by the total number of possible pairs of license plates to get the probability.