Question

A circular dartboard of radius 1 foot is at a distance of 20 feet from you. you throw a dart at it and it hits the dartboard at some point q in the circle. what is the probability that q is closer to the center of the circle than the periphery?

Answer 1

For a better understanding of the solution given here please go through the diagram in the attached file.

The diagram is a representation of the circular dart board with radius 1 foot.

Obviously the dart will be closer to the centre if it falls within half foot of the radius as represented by the region $R_1$. Likewise, the dart will be closer to the periphery if it falls on the region between the periphery/circumference and the half foot radius which is denoted by $R_2$. Thus, we can see that the probability of dart falling in any region on the board is directly related to the area of the region.

Let $P(R_1)$ denote the probability that the dart falls on region, $R_1$, that is closer to the center of the circle than the periphery. We know that bigger the area, of the region $R_1$, bigger will be the probability.

Thus, $P(R_1)\propto\pi(\frac{1}{2})^2$ (because the radius of region $R_1$ is $\frac{1}{2} ft$)

Now, let us assume that the total probability of the dart hitting the board is 1. Then, when the radius is 1 foot, then probability is 1, for an area of $\pi\times (1)^2=\pi$

Therefore, the probability that point q, where the dart falls, is closer to the center of the circle than the periphery is: $\frac{1}{4}$

Answer 2

Answer:

Step-by-step explanation:

by assuming closer to centre of the circle is less than half the radius and closer to periphery is more than half the radius from centre.

Divide the dart into two concentric circles. Radius of smaller is r/2 and bigger is r.

Dart is closer to centre Q lies in smaller circle.

Probability of point Q lying closer to the centre = (Area of Smaller circle)/(Total area of dart board)

= (π × (r/2)2)/(π r2)

= 1/4 = 0.25