Dart board problems

Since there are only 1-point- and 2-point-fields on the board, ths only way to get an odd sum is to have 3 points, so either 1 and then 2 or 2 and then 1.

The area of the dartboard is given by $A_{board} = \pi r^2 = 36 \pi$ . To get the area of the orange regions (for 2 points), we imagine rotating the inner circle by 120° so that the big and a small orange region line up to have area $A_{orange1} = \frac 1 3 A_{board} = 12 \pi$ . The other orange region is one third of the smaller circle with radius 3, so it has an area of $A_{orange2} = \frac 1 3 \pi r^2 = 3 \pi$ . In total, the orange regions have an area of $A_{orange} = A_{orange1} + A_{orange2} = 12 \pi + 3 \pi = 15 \pi$ .

The area of the blue regions is just $A_{blue} = A_{board} - A_{orange} = 36 \pi - 15 \pi = 21 \pi$ .

The probability of an odd sum is $p(1) \cdot p(2) + p(2) \cdot p(1) = 2 p(1) p(2). p(1)=\frac{21 \pi}{36 \pi}=\frac{7}{12}, p(2)=\frac{15 \pi}{36 \pi}=\frac{5}{12}.$

$2 p(1) p(2) = 2 \cdot \frac{7}{12} \cdot\frac{5}{12} = \frac{35}{72}$ .

Therefore, $a+b = 35+72 = \boxed{107}$ .