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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?

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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.

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  • $\begingroup$ Oh! Oof I fail at dis lol xD! Thanks! :P $\endgroup$
    – Max0815
    Mar 13 '19 at 23:39
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    $\begingroup$ No worries, double counting is a common mistake $\endgroup$
    – Shailesh
    Mar 13 '19 at 23:41
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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 $\frac{1}{6}$.

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