<s>[INST] Probability question about choosing digits between 1 and 9 and that their product is a multiple of 3 Two digits between 1 and 9, inclusive, are selected at random. The same digit may be selected twice. What is the probability that their product is a multiple of 3?
I used this logic: if any one of the digits selected is 3, then the product would be a multiple of 3. There are 3 multiples of 3 (3, 6, 9) between 1-9, so $\frac{3}{9}$ = $\frac{1}{3}$, and since the second number can be anything, $\frac{1}{3} * \frac{1}{1} = \frac{1}{3}$. Therefore, my final answer is $\frac{1}{3}$. However, the answer key used complementary counting and got a different answer ($\frac{5}{9}$). What is wrong with my approach?
 [/INST] Total number is: $9\cdot 9=81$ (9 possibilities for the first digit and 9 for the second).
Number of combinations that are not multiple of 3: $6\cdot 6=36$ (6 possibilities for the first digit and 6 for the second), 6 because from the set {1,2,4,5,7,8}.
Number of combinations that are multiple of 3: $81-36=45$.
Probability: $\frac{45}{81}=\frac{5}{9}$.
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