Conditional probabilities based on a drunk man walking on a straight line

Question

Probability of forward step, P(F) = p

Probability of backward step, P(B) = 1-p

(a) Calculate probability that after two steps he will be at the his original place.

(b) Calculate probability that after three steps, he will be one step forward from where he began?

(c) After three steps he managed to move one step ahead, what is the probability his first step was a forward step?

Solve using conditional probability and/or bayes theorem.

For a) is the following correct??

Ways in which the event can take place:

Forward step and backward step

backward step and forward step

P(event)=P(BF) + P(FB) = P(B)P(F) + P(F)P(B) = 2p(1-p)

Answer 1

the random variable that describes you "random walk" is the following

$$X \equiv \begin{cases} 1-p, & \text{if $x=-1$ } \\ p, & \text{if $x=1$} \end{cases}$$

(a) to return at origin after 2 steps, the only 2 cases are the following

• first backward and second forward
• first forward and second backward

The probability is then $2p(1-p)$

(b) similarly, the only 3 cases are the following

• $\{+1;+1;-1\}$

• $\{+1;-1;+1\}$

• $\{-1;+1;+1\}$

each event has a probability of $p^2(1-p)$ hence the probability for (b) is simply $3p^2(1-p)$

(c) Observe that, given the 3 events in (b) there are 2 events that match with your request: First step Forward hence the probability for (c) is $\frac{2}{3}$