Re: probability Message #9 Posted by Karl Schneider on 14 Oct 2004, 2:13 a.m., in response to message #8 by KC
The problem can be approached as a "nested recursive summation" that is not too difficult to program on an RPLbased HP calc (28/48/49 series).
For the 3variable example (let's call it red, green, and blue marbles), the 10 ways to choose three marbles is as follows:
R G B arrangements
3 0 0 1
2 1 0 3
2 0 1 3
1 2 0 3
1 1 1 6
1 0 2 3
0 3 0 1
0 2 1 3
0 1 2 3
0 0 3 1
total: 27
There are (R+G+B)!/(R!G!B!) number of each arrangement
The RPL expression (Sig = Greek letter sigma)
'Sig(R=0,3,Sig(G=0,3R,1))'
can quickly be evaluated as 10.
(NOTE: Knowing the number of red and green marbles, there is only one possible value for the number of blue ones. So, it is not necessary to include it in the summation.)
Neither my 48G nor my 49G could procude the correct answer for the 10variable case, the way I programmed it. THe 48G got stuck in a loop; the 49G seems to have different syntax.
 KS
