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(a)
We can restrict our attention to events in B. What then is the probability that such an event also lies in A? By the Venn diagram we can see that this probability must be
(b)
Now we can restrict our attention to events in . Since the events in lie in both A and B, then the quantity counts such events twice. Thus,


(c)
If we consder tossing a circle of radius a onto the plane, then the expectation value for the number of intersections is 4. The Java applet below (a = 1.76) will let you move a circle of radius a around the plane. You will see that "almost all" circles generate four intersection points. [Can you find the circles that have only three intersection points?]
Created with Cinderella
Recall that E is linear, so that for some real constant r, where l is the length of a curve. Thus we must have that since our circle has length , and so . Now for any needle of length l < a, we know by the defintion of expectation value that the expected number of intersections of our needle with the square grid is given by E(l). [Note that since the probabilities are all equal to zero for n > 2, as our needle can never have three or more intersections.] Thus,
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