In the spinner shown, all circles have the same centre:

- The inner ring is divided into equal sections numbered 1 and 2.
- The middle ring is divided into equal sections numbered 1,2 and 3.
- The outer ring is divided into equal sections numbered $1,2,3$, and 4.

![](https://cdn.mathpix.com/cropped/2024_04_17_d0cbae19e125c7a52478g-3.jpg?height=284&width=293&top_left_y=178&top_left_x=1664)

After the arrow is spun, it lies on one section of each of the three rings and the score is the total of the three numbers in these sections. For example, in the spinner shown, the score is $2+2+3=7$. The arrow is spun once. What is the probability that the score is odd?
We divide the spinner into 12 equal sectors as shown. The score obtained if the arrow lands in each sector is written just outside the arc forming the outer boundary of the sector.

![](https://cdn.mathpix.com/cropped/2024_04_17_698c2a84b151b65f3cc1g-10.jpg?height=610&width=616&top_left_y=1928&top_left_x=814)

The spinner lands in each of these 12 sectors with equal probability, so the probability that the spinner lands in any one of the 12 sectors is $\frac{1}{12}$.

Of the 12 sectors, 10 have an odd score, so the probability of spinning an odd score is $\frac{10}{12}=\frac{5}{6}$.

ANSWER: $\frac{5}{6}$