We'll keep track of the random variable R , which is the distance the dart lands from the center.
Some Questions:
For now let's return to the simple experiment.
With the darts falling at random anywhere in the circle it should seem reasonable that:
In the case of the dart variable R,
{  0 when A <= 0  
F(A) =  A^{2} when 0 < A < 1  
1 when A >= 1 
The probability that the darts would fall in any particular band (called an annulus) formed by concentric circles is also easy from the areas.
The probability that A < R <= B is just F(B)  F(A).
With this analysis it should be clear that the probability that R = A is zero since the circle of radius A is a region in the plane with area zero.
This result can be interpreted as saying that the likelihood of the dart landing on a circle of a given radius is very small. And in an experiment, any specified number A from 0 to 1 is equally likely to occur as the value of R.
Yet the formula above also suggests that the probability that the value of R will lie between 1/8 and 1/4 is not as large as the probability that R will lie between 3/4 and 7/8.
This leads to the concepts of average probability density and point probability density.
The average probability density for an interval [A,B]is the ratio of the probability that R will fall in a certain interval [A,B] to the length of that interval, BA. That is,
The densities for the intervals [1/4,3/8] and [3/4,7/8] illustrate why larger values of R are more likely by measuring the average density of comparable length intervals that contain them.
For the interval [1/4,3/8] we have the density is
while for the intervals [3/4,7/8] the density is
The point probability density of the random variable R at the point A, dF(A), is the limit as B>A of the average probability densities for intervals with endpoints A and B. So
dF(A) = 




This is the key relation between the distribution function F and the probability density function of a random variable.
REMARKS on the DENSITY FUNCTION.
So to find the probability that a random variable is between
A and B we need only integrate its density from A to B.
Using the area interpretation of the definite integral,
the probability that a random variable is between A and B we find the area
of the region bounded by the Xaxis, the graph of the density function
for the random variable, and the lines X=A and X=B.
Let's return to our darts. One problem is to find a
value for A so that the probability that R
<= A is the same as the probability that R > A.
It should seem reasonable that this probability is
1/2.
This number, A, is called the MEDIAN of R.
So...
we want F(A) = A^{2 }= 1/2. Not too hard for
algebra...
Another problem is to find the point with the highest density. This number is called the MODE of the random variable this is the number where for any given small interval length the probability of R being in that interval is highest.
For the darts random variable R, the mode is the A with
the largest value of 2A in the interval from 0 to1,
that is , the mode of R is 1!
First let's note that F(B)F(A) is approximately dF(A)
* (BA)
or dF(B)*(BA) or dF(M)*(BA) where M = (A+B)/2
when B is close to A.
Let's cut the interval from 0 to 1 into n pieces each length 1/n. For example if N = 4. We might consider 16 darts thrown to determine roughly in which of the four sections the darts would fall.















estimated average
= (1/8)(1/16)+(3/8)(3/16)+(5/8)(5/16)+(7/8)(7/16)
= (1+9+25+49)/128
= 84/128
= 21/32
This would be an underestimate for the average. Can you
see why?
THE MEAN MEETS THE RIEMANN INTEGRAL:
Now remember that F(A_{k})  F(A _{k1})
is approximately dF(M_{k})_{ }*( 1/N)
by our earlier remark. So we have then that
to estimate the average value of R theoretically
we could consider S
(M_{k}) *( F(A_{k})  F(A _{k1}))
AHahhh! this last expression is precisely a Riemann Sum that estimates the definite integral
AND...
in general the MEAN of the random variable R must be
the integral of x*dF(x) over the interval for which dF(x) > 0.
The distribution for that variable is the solution to the differential equation
It is an interesting problem to find the value of k here, but that's another story which perhaps you'll find in some further course in calculus.....