Question

If a point is selected at random in the interior of a circle, find the probability that the point is closer to the center than to the circle.

Answer 1

To the circle's circumference?

Answer 2

I don't know that is all the question says

Answer 3

All right, so we have a circle with radius \(r\), and we know the point is closer to the center when it is within a radius of the center of \(\frac{r}{2}\). So, we compute the area of the circle to find that: \[
\pi \left(\frac{r}{2}\right)^2=\pi\frac{r^2}{4}
\]We know the area of the outside part (or half) of the circle is:
\[
\pi r^2-\pi\frac{r^2}{4}=\pi\left(r^2-\frac{r^2}{4}\right)=\pi r^2\left(1-\frac{1}{4}\right)=\pi r^2\frac{3}{4}
\]So, we take the ratio between the areas, and we get a probability of \(\frac{1}{4}\)

Answer 4

Sorry, of \(\frac{1}{3}\).

Answer 5

I do believe that the probability is calculated by:
\[ \ \frac{\text{favorable outcomes}}{\text{total outcomes}} \]
What you were calculating is the odds of choosing the specified point:
\[ \ \frac{\text{favorable outcomes}}{\text{unfavorable outcomes}} \]

Answer 6

Well, yeah, then in the case it would be \(\frac{1}{4}\) as my first post said.

Answer 7

wait i don't get why you subtract and everything. can't u just divide your answer in the first equation by the original pi times r squared

Answer 8

Yes, that is the correct process. You need only calculate the area of the part closest to the center, and then take the ratio closest/total.

Answer 9

^Eeyup, sorry, for some reason or another I thought you were looking for odds as I was typing, rather than probability...