Probability of choosing a number from the set $\{1,2,\ldots,99\}$ that divided by $5$ has the remainder $2$ and is a multiple of $3$

Question

Good evening to everyone. I have to find the probability of choosing a number from the set {1,2...99} that divided by 5 has the remainder 2 and at the same time it's multiple of 3. I know that the probability of an event is the ratio of the number of observations of the event to the total numbers of the observations.The number of observations of the event is $6$ $(17, 32, 47, 62, 77, 92)$ therefore $ \frac{6}{99} = \frac{2}{33} $. But my friend told me that I didn't solve it correctly. How do I solve this problem? Thanks for any response.

Answer 1

Your final answer is correct, but you didn't obtain it correctly.

For a number to have remainder $2$ when divisible by $5$, it must end in a $2$ or a $7$. Of the numbers $2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97$, which are divisible by $3$? Only those whose digits add up to a multiple of $3$. Namely, $12, 27, 42, 57, 72, 87$. Thus, there are $6$, so the probability is $6/99=2/33$.