Calculating Geometric Probability on a Round Table

[Yankel]

Hello all,

I have a question related to geometric probability. I think I solved it, but not sure, would appreciate your opinion.

We are given a round table with a radius of 50cm. At the center of this table there is another circle, with a radius of 10cm. A coin with a radius of 1cm is thrown on the table. Assuming that it landed on the table, what is the probability that the coin (the entire coin) is within the small circle ?

I said that the area of the big circle is

\[2500\pi\]

this is the sample space.

The set of the required event is the points creating the area of the small circle, but going 1cm inside, to allow the entire coin to be inside, so it is:

\[81\pi\]

Therefore the probability is:

\[\frac{81}{2500}\]

Am I correct ?

Thank you !

 

[MarkFL]

I would be inclined to say:

\(\displaystyle P(x)=\frac{\pi\left(10-\dfrac{1}{2}\right)^2}{\pi\left(50-\dfrac{1}{2}\right)^2}=\left(\frac{19}{99}\right)^2\)

 

[Opalg]

Not sure where Mark's $\dfrac12$ is coming from. The question says that the radius (not the diameter!) of the coin is 1cm. So I would give the answer as $\left(\dfrac9{49}\right)^2$. This assumes that the coin has to land so that it is entirely on the table. The OP's answer $\dfrac{81}{2500}$ assumes that it is allowed to land so that it precariously overlaps the edge of the table.

 

[MarkFL]

Yes, I misread radius as diameter...oops. :)