Probability of 2-digit appearing in random 3 digit number? Let's say I have {1, 2} what is the probability of them both appearing in three digit number?
[1x2, 2x1, 12x, 21x, x12, x21] 
I tried the following but is not getting the correct answer
1 - P(Both not appear) - P(Only 1 appear) = 1 - 0.8 * 0.8 * 0.8 - 0.2 * 0.8 * 0.8 *3 = 0.104.
The answer should be 0.054. What is wrong with my work?
EDIT
I manage to work it out
1- P(Three Others)-P(Three 1s)-P(Three 2s)-P(Two 1s,no 2)-P(Two 2s,no 1)-P(One 1,no 2)-P(One 2,no 1)
=1-0.8^3-2*0.1^3-2*3*0.1^2*0.8-2*3*0.1*0.8^2=0.054
 A: What you calculated is the probability of getting a 3-digit number where at least two of the numbers are either a $1$ or a $2$. But note that that would include such numbers as $111$ and $226$, which is not what you want. So, you need to make sure you get at least one $1$ and at least one $2$
To repair this, you can subtract the ones as indicated: you have exactly one $111$ and exactly one $222$.  Now you need to subtract the ones with two $1$'s and something that's neither a $1$ or a $2$ (so there are 8 digits times 3 locations = 24 such numbers) and finally the ones with $2$'s and something that's neither a $1$ or a $2$ (also 24).  Subtracting those $1+1+24+24=50$ patterns should get you from $104$ to $54$ as desired
Or, in terms of probability: the probability of getting $111$ is $0.1\cdot 0.1 \cdot 0.1=0.001$. Likewise, the probability of getting $222$ is $0.001$.  The probability of getting two $1$'s and something that's neither a $1$ or a $2$ is $0.1\cdot0.1\cdot 0.8\cdot3=0.024$, and likewise the probability of getting two $2$'s and something that's neither a $1$ or a $2$ is $0.024$
A: You're off to a good start with the six patterns; now you just have to figure out how many ways there are to fill in the $x$.  To avoid problems of double-counting, I would first count only the cases where $x\neq1,2.$  Then I'd count separately the ways to get two $1$s and a $2$ or two $2$s and a $1$.
I get $.052$ though, not $.054$ are you sure that's supposed to be the answer?
