In an unrestricted random walk the probability that a partic

In an unrestricted random walk, the probability that a particle moves 2 steps to the right is p, and 1 step to the left is q = 1 – p. Xn is the random variable representing the position of the particle to the right of the starting position after n moves.

(a) List all the possible sample paths taken by the particle that lead to

(i) X5 = 1;

(ii) X5 = 2.

(b) Hence express in terms of p and q

(i) P(X5 = 1);

(ii) P(X5 = 2).

Solution

Let a step on right be indicated as \'R\' and that on left be indicated as \'L\'.

At each step, there are 2 possibilities to move ahead. So, for 5 steps, possible number of moves = 2⁵ = 32.

Below are the 32 possible cases and the final destination in each case -

The Final position is nothing but the sum of all the 5 steps it takes where, R = 2, and L = -1. The probability of R in each step is \'p\' and that of L is \'q\'. So, the probability of x steps taken right and y steps taken left will be = p^xq^y. In this way we obtain the probability.

Now, we can answer the question as below -

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(a)

(i) All sample paths taken so that X5 = 1 are as shown below -

ii) There are no path which can return X5 = 2. As it can be seen in the table we created.

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(b)

(i) P(X5 = 1) is the sum of probabilities of all the cases when X5 =1.

So, P(X5 = 1) = 10 q³p².

Because there are 10 possible occurances and each occurance has the same probability of q³p².

Hence, the required probability = P(X5 = 1) = 10 q³p².

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(ii) As there are no possible cases for X5 = 2, its probability is zero.

Hence, P(X5 = 2) = 0.