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 :

$Area = \frac{\pi r^2 \theta}{360}$ , where r is the radius of the spinner.

$Area = \frac{\pi r^2 *80}{360}$

$Area = \frac{2 \pi r^2}{9}$

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 $= \frac{\frac{2}{9}\pi r^2}{\pi r^2}$

Probability $= \frac{\frac{2}{9}}{1}$

Probability $= \frac{2}{9}$

Hence required probability is 2/9.