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.
Best Answer
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.