What is the probability of the dog to reach town A within a day (Word Problem)

Question

A long and narrow railed bridge is situated between town A(to the east) and town B(to the west). A dog is placed at the center of the bridge and it can move about randomly across the bridge. Based on thousands of this same experiment conducted in the past, it is concluded that the probability of any dog placed at the center of the bridge to reach town A is 10% within a day and there is also a 10% chance that any dog placed at the center of the bridge would reach town B within a day. Therefore, the 4 possible outcomes of the experiments are that the dog could reach: Town A, town B, neither or both towns within a day.

What is the probability of the dog to:

(i). Reach town A within a day and,

(ii). Town B must not be reached prior to the occurrence (i) above.

Answer 1

This question does not quite have enough details to answer properly, so I'm going to make some assumptions. Namely, I will assume that the dog reaching town $A$ does not affect the probability of whether or not it reaches town $B$ that same day, and vice versa. Moreover, I am assuming that once a dog reaches a town, it does not go back to it directly afterwards (e.g. the dog reaches town $B$, leaves, then goes back to town $B$.)

Let $A$ be the event that the dog reaches town $A$. Let $B$ the event that it reaches town $B$. The only possible things that could happen are:

1. The dog does not reach either town.
2. The dog reaches $A$ and not $B$.
3. The dog reaches $B$ and not $A$.
4. The dog reaches $A$, then goes to $B$.
5. The dog reaches $B$, then goes to $A$.

What are the probabilities of each of these?

1. $P(\lnot A\land\lnot B)=P(\lnot A)P(\lnot B)=0.9\cdot0.9=0.81$.
2. $P(A\land\lnot B)=P(A)P(\lnot B)=0.1\cdot0.9=0.09$.
3. $P(\lnot A\land B)=P(\lnot A)P(B)=0.9\cdot0.1=0.09$.
4. $\frac12P(A\land B)=\frac12P(A)P(B)=\frac12\cdot0.1\cdot0.1=0.005$.
5. $\frac12P(A\land B)=\frac12P(A)P(B)=\frac12\cdot0.1\cdot0.1=0.005$.

To get 4 and 5, note that the event of reaching both towns, $A\land B$, can happen in two ways, each of equal probability.

Hence, we want to know what is the probability of 2 or 4 happening, which is $0.09+0.005=0.095$.