"riverman" wrote in :

Thus, all the possible positions of A and C equal all the possible
positions of C and A...


But the question as you pose it has nothing to do with all the possible
positions of dart A-- it has to do with one specific position of dart A!
Let's say that A is 5cm away. Then you are looking for p(C5), which has
a value that depends only on the distribution of dart C.

Whatever high probability exists if A is close is
countered by the low probability if A is far.

True before dart A is thrown, but not after dart A is thrown. Now, you
have a real honest to goodness value for dart A.


Thus, all the possible positions of A and C equal all the possible
positions
of C and A....its a combinatoric problem. Specifically because I
DON'T give the position of dart A.

This doesn't mean that you can just ignore the fact that Dart A is stuck
at a precise location in the dartboard, and it's why there isn't enough
info to offer a p-value.

If its any comfort, I'm not making this problem up.

Then the person who did got it wrong.

--
Scott
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