Question

You are the observer of a peculiar individual that enters a room with two stations: each
station has a coin. Suppose that the coin at station A has a 40% chance of landing on heads,
while the coin at station B has a 50% chance of landing on heads. The individual plays the
following, monotonous game: if the coin lands on tails, he will stay at the station; otherwise,
he will move to the other station.
(a) Recall that a state transition diagram is a graphical representation of nodes and
arrows with the probabilities of moving from one station to another to label those
arrows. Draw a state transition diagram for the above scenario.
(b) Find the probability of moving to station B in one coin flip, given that the individual
is at station A.
(c) You decide to keep track of the stations that the individual has visited through a string
of characters, i.e. 'AAB' is the event that the individual started at A, stayed A, and
then moved to B. What is the probability that the string will read 'ABBAB' in the
next 5 plays of the “game.” You may assume that the individual started at A with
probability 1

Answer 1

Find the solution in the images attached.

Note: The movement from one station to other is dependent on the result of the toss and the station at which the toss occurs.The result of a coin toss at either station is not dependent on the result of previous coin toss but the movement of the individual is dependent on the result of the coin toss.

Part b) The question is asking the probability that a person at station A would move to station B in one coin flip. The movement from one station to other is dependent on the result of the toss and the station at which the toss occurs. It is not specified where the person was earlier or what was the result of coin tosses earlier. It is given that the individual is at station A and we are to find the probability of his next movement to station B.

    Probability of any movement from point x to y= Probability that individual is at x * probability of move to y (from x)

    You could simply use the state transition diagram for this question but it is important to understand that it is crucial to set up the individual's initial position (as you will see in part c).

   To answer this question using the state diagram, follow the arrow going from state A to state B (Event = Heads & Probability = 2/5 = 0.4=40%). Imagine this to be the starting of the game and you as the player are at station A, there is a 40% chance that you will move to station B following the coin toss.

Part c) Here we are keeping a track of the player's movement. For 5 tosses there would be 2 ⁵ possibilities and according to that our players movement would also be different: (considering A as initial point)

HHHH ==> A--heads-->B--heads-->A--heads-->B--heads-->A ==> ABABA
HHHT ==> A--heads-->B--heads-->A--heads-->B--tails-->B ==> ABABB
HHTH ==> A--heads-->B--heads-->A--tails-->A--heads-->B ==> ABAAB
HTHH ==> A--heads-->B--tails-->B--heads-->A--heads-->B ==> ABBAB
THHH ==> A--tails-->A--heads-->B--heads-->A--heads-->B ==> AABAB
HHTT, HTHT, THHT and so on...

So, the movement from one station to other is dependent on the result of the toss and the station at which the toss occurs.

Probability of any movement from point x to y= Probability that individual is at x * probability of move to y (from x)
Probability of any movement from point x to y to z= Probability of any movement from point x to y * probability of move to z (from y)
or, we can say,
Probability of any movement from point x to y to z= Probability that individual is at x * probability of move to y(from x) * probability of move to z (from y)
Therefore, Probability to move from A to B to B to A to B = Probability that individual is at A * probability of move to B(from A) * probability of move to B(from B) * probability of move to A(from B) * probability of move to B(from A)