Rescaling a probability I can't ge me head around this.
I know that between 00:00h and 00:30h (i.e. within 30 minutes) a person is with a chance of 90% in room A, 7% in room B and 3% in room C.
Now the task is, to derive a presence profile with a resolution of 10 minutes. How do I rescale thes probabilities, because I think there is the following problem: 
If I draw 3 random numbers, the chance that he's in the room all 30 minutes (with the probabilities kept the same), results in 0.9 * 0.9 * 0.9 --> 0.729 which is lower than the original distribution.
Of course I could rescale the probabilities by $\sqrt[3](0.9)$, which would lead me to the desired result. But I would have to do this with the other possibilities as well, so I'd get approximatly these possibilities:
Room A: 0.965, Room B: 0.412, Room C: 0.311, in sum that is > 1, which doesn't make sense for a probability distribution. Just dividing by the sum of all three isn't the right choice either, I believe.
Is there a way to handle this mathematically correct?
 A: This may or may not be what you mean by resolution 10 minutes, but it might mean that the 30 minute interval is marked off into three 10 minute intervals, and during any 10 minute interval the person is in room A,B, or C with the given probabilities of .9, .07, and .03. At the end of the first 10 minute interval the person could change rooms or remain where he is, with the likelihoods to be in each room during the second 10 minutes the same as for the first 10 minutes. 
With this interpretation, one gets sequences of length three telling where the person was during the three 10 minute intervals. For example (A,B,B) would mean person was in room A first, then switched to B, then stayed in B. That probability would be (.9)(.07)(.07). In all there would be 27 sequences of length three with each entry being one of A,B,C and the probability of each of these can be computed by multiplying the probabilities of the rooms in the given sequence of three rooms. 
That the total of these probabilities for the 27 strings winds up being 1 would then follow, since $(.9+.07+.03)^3=1,$ and the result of "multiplying out" the cube here gives each pattern of probability multiplications for the 27 sequences of length three from the rooms A,B,C.
