3 sided dice with the probability
P(1)=0.5
P(2)=0.3
P(3)=0.2
What is the probability of (1) appear once if I roll the dice 3 times?
using binomial I got
3!/2!(0.5)(0.3)(0.2) = 0.09
but if I use the truth table, I got different result
(221,212,122) (331,313,133) (123,132,213,231,312,321)
3(0.5)(0.3)^2 + 3(0.5)(0.2)^2 + 6(0.5)(0.3)(0.2) = 0.375
I'm pretty sure the truth table is correct. but how to translate that to binomial formula?
If you use the binomial law the probability to get 1 once is
$$\mathbb{P}(\textrm{1 appears exactly once}) = {3 \choose 1}0.5^1 (0.3+0.2)^{3-1} = 3 \cdot 0.5 \cdot 0.5^2 = 0.375$$
And you get the same result as the truth table.
The way you applied the binomial law is indeed incorrect. Recall that the formula is $P(X=k)={n \choose k}p^k(1-p)^{n-k}$ if you repeat $n$ times an experiment and you want $k$ successes (with the probability of a success being $p$). $X$ is the random variable counting the number of successes.
In your case $p=0.5$, $k=1$ and $n=3$.