A circular dial is divided into 10 equal parts from the origin toward the circumference. The parts are labeled clockwise from 1 to 28 in increments of 3. Beginning directly to the right, they are marked as 1, 4, 7, 10, 13, 16, 19, 22, 25, and 28. A needle is fixed vertically at the center, with the arrow pointing directly upward to the line between 22 and 25 at the top center of the dial.

Question

A circular dial is divided into 10 equal parts from the origin toward the circumference. The parts are labeled clockwise from 1 to 28 in increments of 3. Beginning directly to the right, they are marked as 1, 4, 7, 10, 13, 16, 19, 22, 25, and 28. A needle is fixed vertically at the center, with the arrow pointing directly upward to the line between 22 and 25 at the top center of the dial.

If this spinner is to be spun once, how likely will it stop on a multiple of 9?

Answer 1

To determine the probability of the spinner stopping on a multiple of 9, we need to first find out how many multiples of 9 are on the dial.

The multiples of 9 on the dial are 9, 18, and 27. So there are 3 multiples of 9 out of a total of 10 possible outcomes.

Therefore, the probability of the spinner stopping on a multiple of 9 is 3/10 or 0.3. So, the likelihood of the spinner stopping on a multiple of 9 is 30%.

Answer 2

So is it certain, unlikely, impossible, likely?

Answer 3

In terms of likelihood, a probability of 0.3 (or 30%) indicates that it is likely for the spinner to stop on a multiple of 9. So, it is likely for the spinner to stop on a multiple of 9 when spun once.