Question

A spinner with 3 sectors is shown in the figure below. The measures of the central angles of the red, white, and blue sectors, in that order, are in the ratio 2:3:4. If the spinner is fair, what is the probability that the arrow stops in the red sector?

(Note: Assume the arrow stops in a sector and not on a line.)

112
29
13
49
59

Answer 1

Given that a spinner with 3 sectors. The measures of the central angles of the red, white, and blue sectors, in that order, are in the ratio 2:3:4.

So we can write:

Measure of the central angle of the red sector = 2x

Measure of the central angle of the white sector = 3x

Measure of the central angle of the blue sector = 4x

We know that sum of central angle of the circle is 360 degree so the sum of above three angles will also be 360 degree

2x+3x+4x=360

9x=360

x=40

Then measure of the central angle of the red sector = 2x=2*40= 80 degree

Measure of the central angle of the white sector = 3x=3*40=120 degree

Measure of the central angle of the blue sector = 4x=4*40=160 degree

We have to find the probability for red sector so first we need to find are of the red sector which is given by formula :

`Area = (\pi r^2 \theta)/(360)` , where r is the radius of the spinner.

`Area = (\pi r^2 *80)/(360)`

`Area = (2 \pi r^2)/(9)`

Area of the spinner is circular so that is given by

Area of spinner
`= \pi r^2`

Now the probability that the arrow stops in the red sector is given by ratio or area of red sector and the area of spinner

Probability
`= ((2)/(9)\pi r^2)/(\pi r^2)`

Probability
`= ((2)/(9))/(1)`

Probability
`= (2)/(9)`

Hence required probability is 2/9.