Question

Three balls are divided between two containers. During each period a ball is randomly chosen and switched to the other container. (a) Model this situation as a stochastic process. (b) Find the transition matrix Container 1 contains one ball at present (c) After two periods, what is the probability that Container 1 contains 2 balls? (d) After two periods, what is the probability that Container 2 contains 2 balls? (e) After three periods, what is the probability that Container 1 contains 3 balls?

Answer 1

(a)

Let S0, S1, S2, S3 be the states that the Container 1 contains 0, 1, 2, 3 balls respectively.

The given problem can be modeled as Markov chain as the transiton probability from state Si to state Sj (for i,j = 0, 1, 2, 3) depends only on the current state (that is number of balls in container 1 at present).

The transiton probability from state S0 to state S1 is 1. Because all balls are in container 2 and a ball in that container will be chisen and moved to container 1.

The transiton probability from state S1 to state S0 is 1/3. Because, the probability that ball from container 1 is chosen is 1/3.

The transiton probability from state S1 to state S2 is 2/3. Because, the probability that ball from container 2 is chosen is 2/3.

The transiton probability from state S2 to state S1 is 2/3. Because, the probability that ball from container 1 is chosen is 2/3.

The transiton probability from state S2 to state S3 is 1/3. Because, the probability that ball from container 2 is chosen is 1/3.

The transiton probability from state S3 to state S2 is 1. Because, the probability that ball from container 1 is chosen is 1 as it contains all the balls.

All other transition probabilities are 0.

(b)

The transiton probability matrix is below where each cell (i,j) denoted the transition probability from State Si to State Sj.

(c)

Let Xn denotes the state after n steps.

The probability that Container 1 contains 2 balls after two periods = P(X2 = S2 | X0 = S1) = 0

as there are no transitions from S1 to S2 in two steps.

(d)

The probability that Container 2 contains 2 balls after two periods = probability that Container 1 contains 1 ball after two periods

= P(X2 = S1 | X0 = S1) = P(X2 = S1 , X1 = S0, X0 = S1) + P(X2 = S1 , X1 = S2, X0 = S1)

= (1/3) * 1 + (2/3) * (2/3) = (1/3) + (4/9)

= 7/9

(e)

Probability that Container 1 contains 3 balls after 3 periods = P(X3 = S3 | X0 = S1) = 0

as there are no transitions from S1 to S3 in three steps.