



 
Hi Anagha, I am going to draw the circle with its center at the origin $O$ in the Cartesian plane. I am going to identify a point $P$ on the circle by the measure $t$ of the angle between positive Xaxis and the line segment from $P$ to $O,$ measured counterclockwise from the positive Xaxis. The problem is that $P$ is an unknown point on the circumference of the circle and there is a spinner attached at $O.$ Spin the spinner. What is the probability that it stops, pointing at $P?$ Suppose the probability is $Pr.$ Start by dividing the circumference of the circle with the Xaxis. The two segments of the circumference are specified by \[0^o \le t < 180^o \mbox{ and } 180^0 \le t < 360^{o}.\] These segments are the same size and hence the probability that the spinner points at the segment containing $P$ is $\large \frac{1}{2}.$ Thus $Pr \le \large \frac{1}{2}.$ Further subdivide the circumference by adding the Yaxis to obtain the four segments \[0^o \le t < 90^{o}, 90^o \le t < 180^{o}, 180^o \le t < 120^o\mbox{ and } 270^0 \le t < 360^{o}.\] These segments are the same size and hence the probability that the spinner points at the segment containing $P$ is $\large \frac{1}{4} = \large \frac{1}{2^2}.$ Thus $Pr \le \large \frac{1}{2^2}.$ Further subdivide the circumference by adding the lines through the origin at $45^o$ and $135^{o}.$ This subdivides the circumference into $2^3 = 8$ segments and hence $Pr \le \large \frac{1}{2^3}.$ You can continue this process indefinitely, each time doubling the number of segments from the previous step. Hence at the $n^{th}$ step we have that $Pr \le \large \frac{1}{2^n}.$ We also know that the probability of any event is between 0 and 1 and hence, for all positive integers $n,$ \[ 0 \le Pr \le \frac{1}{2^n}.\] As $n$ approaches infinity the sequence \[\frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3}, \cdot \cdot \cdot, \frac{1}{2^n}, \cdot \cdot \cdot\] approaches zero and hence we are forced to conclude that $Pr = 0.$ I hope this helps, Harley  


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