ME. M. W. CEOFTON ON THE THEOET OF LOCAL PEOBABILITY. 
195 
18, The following problems relate to a circular boundary: — 
1. A random point falls within a given circle and a random straight line is drawn 
across the circle ; to find the chance of the line passing within a given distance of the 
point. 
As the general solution is somewhat complicated, I will take the particular case 
where the given distance is the radius of the circle, which will serve equally well as an 
example of the application of the foregoing principles. 
Let C be the centre of the given circle, P any position of the random point, r the 
radius of the given circle ; draw an equal circle with P as centre ; then 
the number of random lines meeting the given circle and passing within 
a distance r of the point P, is the same as the number of random lines 
cutting both circles ; this number is measured (art. 6) by the excess of 
the two circumferences over an endless band wrapped round them ; that 
is, putting CP=g>, 
27 rr — 2g>. 
If dS be an element of the surface at P, the sum of the favourable cases will be 
F=jJ(27rr-2 ? )^S=2jVr- f ).2T^ f ; 
.-. F=(7T— 1)2^. 
But the whole number of cases is 27rr X vr 2 ; hence the required chance is 
2 
P = 1 -si- 
Fig. 12. 
I will give another solution of this problem : — Let AB be a position of the 
line; take MN=r, then all the favourable positions of the random 
point are within the segment ETIF ; the number of favourable points 
is therefore 
r 2 {%—$ -f- sin <p cos <p). 
We have to multiply this by the differential of CM, and integrate 
from CM=r to CM=0, which will give the favourable combinations 
for all random lines parallel to AB, passing between C and H; 
doubling this, we have the result for all lines parallel to AB ; that is, 
F 0 =2r 2 J (tt— p + sin <p cos <p)d.r coscp 
= 2r 3 J* (tt— (p + sin <p cos <p) sin<p<Z<p; 
F 0 =2r 3 (7r— 
Now if the system of lines parallel to AB revolve through two right angles, we have 
for the measure of the whole number of favourable cases 
F=2*r 3 (7r-f); 
