In problems like this (with small numbers and just two independent variables), I find it often helps to draw a table and mark the cases being looked for.
The product would be a multiple of 3 if either number is 3. I've marked those cases with the xs in this pretty little ASCII art table: 123456789 1 ..x..x..x 2 ..x..x..x 3 xxXxxXxxX 4 ..x..x..x 5 ..x..x..x 6 xxXxxXxxX 7 ..x..x..x 8 ..x..x..x 9 xxXxxXxxXYou can eyeball count that as 6 lines of 9 xs, except that would double-count the ones where the lines intersect (uppercase Xs), so reduce 9 for that. Of course you could count them in some other way to verify. That gives $$ P(\mathrm{product\ is\ multiple\ of\ 3}) = \frac{6 \cdot 9 - 9}{9 \cdot 9} = \frac{5 \cdot 9}{9 \cdot 9} = \frac{45}{81} = \frac{5}{9} $$
That first part is right. Just that I think in the follow-up you missed that the first number can also be anything if the second satisfies the condition. Listing the pairs explicitly helps with that, and the table also exposes the overlapping cases. Practice problem: What's the probability if we change the problem so that the two digits can't be the same? (It's not the same, by my count.) |