# Markov Chains and absorption probabilites

 P: 19 Could someone please help me with this question? A single-celled organism contains N particles, some of which are of type A, the others of type B . The cell is said to be in state i where 0<=i<=N if it contains exactly i particles of type A. Daughter cells are formed by cell division, but rst each particle replicates itself; the daughter cell inherits N particles chosen at random from the 2i particles of type A and 2N-2i of type B in the parent cell. Find the absorption probabilities and expected times to absorption for the case N = 3. I so far have that the absorbing states are i=0, i=3 but have no idea where to go from there
 Mentor P: 12,113 For N=3, you can calculate the transition matrix manually. Many entries are 0, and some others follow from symmetry, so you just need 2 interesting entries.
 P: 19 how do i calculate the entries though, thats where i'm stuck at the moment, i know of course the lines for starting in state 0 and 3, but have no clue about 1 or 2, once i know that the rest of the question becomes fairly trivial, could you push me in the right direction?
 Mentor P: 12,113 Markov Chains and absorption probabilites i=1 leads to AABBBB in the cell before splitting. If you randomly pick 3 of them, what is the probability of getting 0 (,1,2,3) times A?
 P: 19 oh is that standard binomial? so probability of going from state 1 to 0 would be (2/3)^3 which is 8/27 then do the same for the other states? or am i missing something?
 P: 19 i really don't understand the probabilities of getting to the other states, do i not need to also consider what the other cell will contain or is that irrelevant?
 P: 19 I think i finally get it, so probability of 0 A's is equal to (2/3)*(3/5)*(1/2) which is the probability of selecting a B each time Then follow the same method for 1 A taking into account whether you chose the A first, second or third? I hope thats right
 Mentor P: 12,113 That is correct.
 P: 19 Thanks for the help
 P: 2 I also have difficulty in this question.I have calculated that the probabilty of getting 0 'a's is 1/5...probability of getting 1 'a' is 3/5 and probability of getting 2 'a' is 1/5. What is the transition matrix and what are the absorption probabilities? thank you
HW Helper
Thanks
P: 5,204
 Quote by macca1994 I think i finally get it, so probability of 0 A's is equal to (2/3)*(3/5)*(1/2) which is the probability of selecting a B each time Then follow the same method for 1 A taking into account whether you chose the A first, second or third? I hope thats right
If you start with AABBBB and pick three at random, you are looking at the "hypergeometric distribution", which is the probability for choosing k of type A from a population of 2A and 4B, when you choose three altogether. See, eg., http://en.wikipedia.org/wiki/Hyperge...c_distribution or http://mathworld.wolfram.com/Hyperge...tribution.html .
 P: 2 thank you..i have already found the probabilities but how to find the transition matrix please?