Question 705062

At $t = 0$, the position is $x = 0$. At $t = 1$, the position is $x = 1$. This means the first step (Step 1) taken was towards the RIGHT.

We need to find the probability that the walker is at $x = 1$ at $t = 3$. Since the walker is already at $x = 1$ at $t = 1$, the net displacement over the next two steps ($t = 1 \to 2$ and $t = 2 \to 3$) must be zero. This is only possible if the walker takes one step RIGHT and one step LEFT (in any order).

There are two possible sequences of moves for Step 2 and Step 3:

Sequence 1: Moves RIGHT then LEFT (R $\to$ R $\to$ L)

• Step 2 ($t = 1$ to $t = 2$): The walker moves RIGHT. Since the previous step (Step 1) was also RIGHT, the walker repeats its direction. Probability = $p$ (Position becomes $x = 2$)
• Step 3 ($t = 2$ to $t = 3$): The walker moves LEFT. Since the previous step (Step 2) was RIGHT, the walker changes its direction. Probability = $1 - p$ (Position becomes $x = 1$)
• Probability of Sequence 1 = $p(1 - p)$

Sequence 2: Moves LEFT then RIGHT (R $\to$ L $\to$ R)

• Step 2 ($t = 1$ to $t = 2$): The walker moves LEFT. Since the previous step (Step 1) was RIGHT, the walker changes its direction. Probability = $1 - p$ (Position becomes $x = 0$)
• Step 3 ($t = 2$ to $t = 3$): The walker moves RIGHT. Since the previous step (Step 2) was LEFT, the walker changes its direction again. Probability = $1 - p$ (Position becomes $x = 1$)
• Probability of Sequence 2 = $(1 - p)(1 - p) = (1 - p)^2$

Total Probability: The total probability of being at $x = 1$ at $t = 3$ is the sum of the probabilities of these two mutually exclusive sequences: Total Probability = $\text{P(Sequence 1)} + \text{P(Sequence 2)}$ Total Probability = $p(1 - p) + (1 - p)^2$

Taking $(1 - p)$ common: Total Probability = $(1 - p) [p + (1 - p)]$ Total Probability = $(1 - p) [1]$ Total Probability = $1 - p$

Final Answer: $1 - p$