What is the probability that the total score after throwing darts is divisible by $3$. 
To play a game of darts Michael throws three darts at the dart board shown. The number of points $(1,$ $5$ or $10)$ for each of the three regions is indicated. His score is the sum of the points for the three darts. If the radii of the three concentric circles are $1,$ $2$ and $3$ units, and each dart Michael throws hits this dart board at random, what is the probability that his score is evenly divisible by $3?$ Express your answer as a common fraction.





After taking the values modulo $3$, we have $1, 2, 1$. I am pretty sure that the only way we can get divisible by $3$ in this problem is if we have modulos $1, 1, 1$ or $2, 2, 2$ for the darts. This means that the probability is ${\left(\dfrac23\right)}^3+{\left(\dfrac13\right)}^3=\dfrac13$.
I feel as if I am missing something, or am I correct?
Thanks!

EDIT: "At random" means that the likelihood of a dart landing in a region is the total area of that region divided by the total area of the dart-board.
 A: Suppose probability corresponding to modulo $1$ is $p$, then probability correponding to modulo $2$ is $1-p$.
Hence it should be $p^3+(1-p)^3=p^3 + (1-p)^3.$
The radius are $r_1=1, r_2=2, r_3=3$ respectively and the probability is proportional to the area, then
$$1-p=\frac{\pi r_2^2 - \pi r_1^2}{\pi r_3^2}=\frac{r_2^2-r_1^2}{r_3^2}.$$
Since $r_2=2r_1$ and $r_3=3r_1$, then $$1-p=\frac{4-1}{9}=\frac13$$
While the numerical value coincides, you should illustrates that your probability is computed based on the area.
A: The areas $A_1,A_5,A_{10}$ areas are
$$\begin{cases}
A_{10}: &\pi r^2 = pi\\
A_5: &4\pi-\pi=3\pi\\
A_1: &9\pi-4\pi=5\pi
\end{cases}$$
Out of $(1+3+5)\pi=9\pi$, the probabilities to get:
$$\begin{cases}
P(10) &= 1/9\\
P(5) &= 3/9\\
P(1) &= 5/9
\end{cases}$$
Any of the $27$ possible configurations $(a,b,c)$ has the probability
$$\dfrac{p}{27\cdot 3^3 = 3^6}$$
To reach a divisible by 3 score, there is

*

*only one way to make either $(10,10,10)$, $(5,5,5)$ or $(1,1,1)$ of respective probabilities $$\dfrac1{3^6}, \dfrac{27}{3^6}, \dfrac{125}{3^6}$$which sum is $\dfrac{153}{3^6}$

*three ways to make each of $(1,1,10)$ or $(1,10,10)$,
$$3\dfrac{25+5}{3^6}$$
Summing the whole gives
$$\bbox[5px,border:2px solid #ca9]{\dfrac{243}{3^6}=\dfrac{3^5}{3^6}=\dfrac13}$$
