# Finding probability that two chosen numbers have a multiple of nine.

If two numbers will be randomly chosen without replacement from $$\{3, 4, 5, 6\}$$, what is the probability that their product will be a multiple of 9?

What I did: To be a multiple of 9, a number needs to have 2 threes. The only possible choice in the given set is choosing 3 and 6. Choosing either 3 or 6 has a probability of $$\frac{1}{4}$$, and choosing the other number that's left has a probability of $$\frac{1}{3}$$ which gives $$\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12}$$, but this is wrong. What did I do wrong?

## 2 Answers

You counted the case choosing $$3$$ and then $$6$$ and choosing $$6$$ and then $$3$$ as $$2$$ different cases. So the answer should be $$\frac{1}{6}$$. The order does not matter as long as you choose $$3$$ and $$6$$.

Alternatively, number of elements in your sample space is $$4\choose 2$$, which is $$6$$, with only $$1$$ event of interest.

• Oh! Oof I fail at dis lol xD! Thanks! :P Mar 13 '19 at 23:39
• No worries, double counting is a common mistake Mar 13 '19 at 23:41

The first choice has to be a $$3$$ or a $$6$$, with probability $$\frac{1}{2}$$. The other number is chosen with a probability $$\frac{1}{3}$$, net probability is $$\frac{1}{6}$$.