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Problem 218

Combinatorics Difficulty 1.6 Multiple choice CEMC Gauss (Grade 7) · Canada · 2024

A circular spinner is divided into $12$ identical unshaded sections and $3$ identical shaded sections, as
shown.

Each unshaded section is $3$ times
the size of each shaded section. An arrow is attached to the centre of
the spinner. The arrow is spun once. What is the probability that the
arrow stops in a shaded section?

Pick one

Official solution

Since each unshaded section is $3$ times the size of each shaded section,
then together the size of $3$ shaded
sections is equal to the size of $1$
unshaded section.

Thus, the combined size of all sections is equal to $12+1=13$ unshaded sections.

We can now imagine the spinner as having $13$ equal sized sections of which $1$ is shaded.

The probability that the arrow stops in a shaded section is equal to the
fraction of the spinner’s area comprised of shaded sections, which is
$\frac{1}{13}$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.