What is the probability that the total score after throwing darts is divisible by $3$.

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.

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 $$\bbox[5px,border:2px solid #ca9]{\dfrac{243}{3^6}=\dfrac{3^5}{3^6}=\dfrac13}$$