How to solve the following probability word problem? A paratrooper is jumping from a plane. Based on wind conditions, we know that he will land withing the large circular region. The training drill requires the paratrooper to land on the landing pad. Find the probability that the paratrooper doesn't meet the training drill requirement if the circumference of the larger circle is 628 ft and the landing pad has a diameter of 40 ft 
There was a picture attached to the problem so here it is:

So, to solve the following problem I first wanted to find the radius of the Big circle to calculate its Area. 
I did 
$$2πr = 628 $$
$$ πr = 628/2 = 314$$
$$r=314/π$$
$$ R = 99.9493$$
After finding this, I tried to find the area of the big circle 
$$ (Area)(of)(big)(circle) = π(99.9493)^2$$
$$ (Area)(of)(big)(circle) = 31384.08154$$
Now I went on and tried to find the Area of the small Circle 
$$ (Area)(of)(small)(circle) = π(20)^2$$
$$ (Area)(of)(small)(circle) = 1256.637061$$
I subtracted the Area of big circle and the Area of the small circle 
$$ 31384.08154 - 1256.637061 = 30127.44448 $$
Now as it was asking for the probability of the paratrooper landing on the big circle I put the subtracted area of the big circle over the total area of the big circle to get 
$$ 30127.4448/31384.08154 $$ 
But when I divide that I get .95995, and when I try to convert that to fractions, I can't for some reason in my calculator, This made me think that I solved it the wrong way. 
If I did it wrong, How do I solve it? 
 A: Your math was fine until the very end, when you decided the answer has to come out to be a nice fraction.  Indeed, if you were given the circumference of the large circle as $200\pi$ your answer would be $\frac{19}{20}$ but $628$, alas, is not a simple multiple of $\pi$.
The problem, on the other hand, is deeply flawed.  It becomes valid if it adds that the probability of landing point is uniformly distributed over the whole potential landing zone.  Generally, that won't be the case.
A: Your approach is generally correct (although I do not understand why you would want to subtract the two areas), but it is always a good idea inserting the actual numbers as the very last thing you do.
Using big letters to denote the quantities for the big circle (and small for the small circle), we have 
$$C=2\pi R $$
$$\implies$$
$$A=\pi\left(\frac{C}{2\pi}\right)^2 \quad\text{and}\quad  a=\pi\left(\frac{d}{2}\right)^2$$
$$\implies$$
$$P=\frac{a}{A}=\left(\frac{\pi d}{C^2}\right)^2=\frac{\pi^2 40^2}{628^2}\approx 0.04.$$
The probability you're after is then $1-P$.
