So in other words, if X denotes the number of 1's rolled in 10 attempts, and Y denotes the number of 2's rolled in 10 attempts, what is the probability that X=Y=1?
I know the probability of rolling any particular number exactly once in 10 attempts is
10C1 * (1/6) * (5/6)^9
And I was able to calculate in an earlier problem the probability that X=Y=0 (I got (4/6)^10).
But I am unsure how to calculate this when we need two numbers to be rolled exactly once.
The probability of getting a $1$ and $2$ on $2$ rolls is $\left(\frac{1}{6}\right)^{2}$
There are $10 \times 9 = 90$ ways to choose the $2$ positions where $1$ and $2$ will be rolled out of the $10$ rolls.
The probability of getting a $3, 4, 5$ or $6$ on all of the other $8$ rolls is $\left(\frac{4}{6}\right)^{8} = \left(\frac{2}{3}\right)^{8}$.
The final probability hence is
$$90 \times \left(\frac{1}{6}\right)^{2}\times \left(\frac{2}{3}\right)^{8} = \boxed{\frac{640}{6561}}$$