Question

You are given the following transition probability graph:

a) If the process begins in state 1, what is the probability that absorption will occur after exactly five steps (i.e., the absorbing step will be reached on the fifth step)?

b) If the process begins in state 2, what is the probability that absorption will occur in six steps or fewer?

Answer 1

ANSWER:

A)

Starting from 1 the only paths we can take to get to 3 in excatly 5 steps are

121213,121013,101213,101013,100013.Denoted in order of the states visited.

Using the trasformation probabilities to find the probability of each path

P(121213)=P(12)P(21)P(12)P(21)P(13)=1/(2*1*2*1*4*)=1/16

P(121013)=1/(2*1*4*2*4)=1/64

P(101213)=1/(4*2*2*1*4)=1/64

P(101013)=1/(4*2*4*2*4)=1/256

P(100013)=1/(4*2*2*2*4)=1/128

P(1 to 3 in 5 steps)=sum of probabilties of each path=1/16+1/64+1/64+1/256+1/128=27/256

P(1 to 3 in 5 steps)=sum of probabilties of each path=1/16+1/64+1/64+1/256+1/128=27/256

B)

paths that go from 2 to 3 in 2 steps->213

paths that go from 2 to 3 in 3 steps->None

paths that go from 2 to 3 in 4 steps->21213,21043

paths that go from 2 to 3 in 5 steps->210043

paths that go from 2 to 3 in 6 steps->2121213,2121013,2101213,2101013,2100013

It is easier to see this since all paths satsrt with 21 and end with 13,we just loook for the paths that take you from 1 to 1 in 2 steps.

P(2 steps)=P(213)=1/(1*4)=1/4

P(4 steps)=P(21213)+P(21013)=1/(1*2*1*4)+1/(1*4*2*4)=1/16+1/32=3/32

P(5 steps)=P(210013)=1/(1*4*2*2*4)=1/64

P(6 steps)=P(12)P(1 to 3 in 5 steps)=27/256

P(6 or less steps)119/256.