A random walker takes a step of unit length towards right or left at any discrete time step. Starting from x = 0 at time t = 0, it goes right to reach x = 1 at t = 1. Hereafter if it repeats the direction taken in the previous step with probability p, the probability that it is again at x = 1 at t = 3 is

Correct option is A

Solution:

To determine the probability that the random walker is again at x=1 at time t=3, we need to consider the possible sequences of steps it can take after the first step. The random walker starts at x=0 at time t=0 and moves to x=1at time t = 1. From x = 1at time t = 1, the random walker can take the following steps:

1. Right, Left, Right (R, L, R)
2. Right, Right, Left (R, R, L)
3. Left, Right, Right (L, R, R)
4. Left, Left, Right (L, L, R)

However, since the random walker repeats the direction taken in the previous step with probability p, and changes direction with probability 1−p, we need to consider the probabilities of each sequence.

Let's analyze each sequence:

1. Sequence R, L, R:

   • The probability of the first step being Right is 1 (since it is given).
   • The probability of the second step being Left is 1−p (since it changes direction).
   • The probability of the third step being Right is 1−p (since it changes direction).
   • Therefore, the total probability for this sequence is 1×(1−p)×(1−p)=(1−p)².

2. Sequence R, R, L:

   • The probability of the first step being Right is 1.
   • The probability of the second step being Right is p (since it repeats direction).
   • The probability of the third step being Left is 1−p (since it changes direction).
   • Therefore, the total probability for this sequence is 1×p×(1−p)=p(1−p).

3. Sequence L, R, R:

   • The probability of the first step being Right is 1.
   • The probability of the second step being Left is 1−p (since it changes direction).
   • The probability of the third step being Right is p (since it repeats direction).
   • Therefore, the total probability for this sequence is 1×(1−p)×p = p(1−p).

4. Sequence L, L, R:

   • The probability of the first step being Right is 1.
   • The probability of the second step being Left is 1−p (since it changes direction).
   • The probability of the third step being Left is p (since it repeats direction).
   • The probability of the fourth step being Right is 1−p (since it changes direction).
   • Therefore, the total probability for this sequence is 1×(1−p)×p×(1−p)=p(1−p)².

However, we only need to consider the sequences that bring the random walker back tox=1at timet=3. The only sequences that satisfy this condition are R, L, R and R, R, L.

Therefore, the total probability is the sum of the probabilities of these two sequences:

(1−p)² + p(1−p) = (1−p)(1−p+p)  = (1−p)×1 = 1−p

Thus, the probability that the random walker is again at x = 1 at time  t = 3 is 1−p .