# Problem 8 from the Spring 2018 Mathcounts

### Solution

If the probability of hitting a region is proportional to its area, then the odds of hitting one in the outer ring to those of hitting one in the middle ring to those in the central circle are $5:3:1.$

Summing up the probabilities for each number we get $\displaystyle P(1)=\frac{10}{72},$ $\displaystyle P(2)=\frac{14}{72},$ $\displaystyle P(3)=\frac{12}{72},$ $\displaystyle P(4)=\frac{14}{72},$ $\displaystyle P(5)=\frac{10}{72},$ $\displaystyle P(6)=\frac{10}{12}.$

As the dice is fair, the probability of coincidence is proportional to th probability of hitting a number on the dartboard. $2$ and $4$ have the highest probabilities of being hit so that they are most likely to hit the outcome of the die roll.

For the first question, the distribution on the dartboard is inconsequential. Once a number was hit, there is the same probability of $\displaystyle \frac{1}{6}$ that it will be matched by the outcome of a die roll.

### Acknowledgment

This problem is based on problem 8 from the Spring 2018 Mathcount competition.