<s>[INST] Probability that two plates with numbers ranging from $0001-9999$ have at least two common digits in them. I would calculate it as this $\frac{10^4-9^4-9^4+8^4}{10^4}$. But it may be incorrect Because I summing the $10^4-9^4$ and $9^4-8^4$. Is it the correct? 
 [/INST] Let's consider "common digits" to mean the digit appears in both plate numbers in any order. Let's find the probability that the two plates have no common digit.
If the first plate has one distinct digit, this can happen in any of 9 ways ($0000$ is not possible), so there are $9^4-1$ ways to choose the numbers for the second plate.
If the first plate has two distinct digits, there are 14 possible orders for the two digits: $\dfrac{4!}{3!1!}+\dfrac{4!}{2!2!}+\dfrac{4!}{1!3!}$. There are $\dbinom{10}{2}$ ways to choose the two digits. There are $8^4$ ways to choose a four digit number for the second plate without those two digits. We have to subtract off the number of ways to choose the first plate with no zeros and the second plate with all zeros, since that is a forbidden number: $14\dbinom{9}{2}$.
If the first plate has three distinct digits, there are $\dfrac{4!}{2!1!1!}+\dfrac{4!}{1!2!1!}+\dfrac{4!}{1!1!2!} = 36$ ways to order the numbers, and $\dbinom{10}{3}$ ways to choose them. Then there are $7^4$ ways to choose the number for the second plate without any of the three digits from the first plate. However, this overcounts when the second plate is $0000$, so we have to subtract $36\dbinom{9}{3}$.
Finally, if the first plate has four distinct digits: $\dbinom{10}{4}4!6^4-\dbinom{9}{4}4!$
Complement of the total probability that there are no common digits (which means at least one shared digit):
$$1-\dfrac{9(9^4-1)+14\dbinom{10}{2}8^4-14 \dbinom{9}{2}+36\dbinom{10}{3}7^4-36\dbinom{9}{3}+24\dbinom{10}{4}6^4-24\dbinom{9}{4}}{(10^4-1)^2} = \dfrac{8,937,985}{11,108,889} \approx 80\%$$
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