What is the probability that a two digit number is selected at random will be a multiple of 3 and not a multiple of 5?

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yash mittal 2 years ago

Two digits between 1 and 9, inclusive, are selected at random. The same digit may be selected twice.

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.

What is the probability that their product is a multiple of 3?

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 xxXxxXxxX

You 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 $$ P(\mathrm{product\ is\ multiple\ of\ 3}) = \frac{6 \cdot 9 - 9}{9 \cdot 9} = \frac{5 \cdot 9}{9 \cdot 9} = \frac{45}{81} = \frac{5}{9} $$

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 3/9 = 1/3, and since the second number can be anything [...]

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.)

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