# Four towns are connected by five roads

Question:

Four towns are connected by five roads. Imagine a rhombus where the left, up, right, and down corners are $A, B, D$, and $C$ respectively; and connect the corners B and C. Each road is blocked by snow independently with probability $p$. What is the probability that you can drive between the towns $A$ and $D$?

I will explain how I understand the first part of the solution.

More importantly, I am asking for someone to explain the solution which is below.

This question is taken from "Probability and Random Variables, a Beginner's guide" by David Stirzaker, pages $63-64$, which shows the following solution:

Let $R$ be the event that the road BC is open and Q the event that you can drive from $A$ to $D$.

$R'$ means the complement of event $R$.

$$P(Q)=P(Q|R) \cdot (1−p)+P(Q|R′)\cdot p$$

$$P(Q)=(1−p^2)^2\cdot (1−p)+(1−(1−(1−p)^2)^2)\cdot p$$

I now will explain my understanding of the first part of the above solution:

(1) $p^2$ is the probability of having two consecutive roads (such as $AB$ and $BD$; or $AC$ and $CD$; or $AB$ and $BC$; or $AC$ and $CB$) blocked. Note that in the case of $AB$ and $BC$, or $AC$ and $CB$, the town $D$ was not reached.

(2) $1-p^2$ is the probability that two consecutive roads are not blocked, and $(1-p^2)^2$ is the probability that two pairs of consecutive roads are not blocked, such as $AB$ and $BD$; or $AC$ and $CD$.

(3) $(1-p^2)^2 \cdot (1-p)$ is the probability that two consecutive roads are not blocked plus road $BC$ not being blocked. So that means the routes $AB$ and $BC$ and $CD$; or $AC$ and $CB$ and $BD$. The problem with this solution is that, apparently, it ignores the routes $AB$ and $BD$; or $AC$ and $CD$.

That was my attempt at understanding the first term of the solution. I invite an explanation of the book's solution (both, the first and second terms of the solution) and, ideally, show step by step how such solution was found, hopefully showing equations with unions and intersections of events.

## 1 Answer

For simplicity, I will write $O_\text{road}$ to indicate the event that a particular road is open, for example $O_{AB}$ denotes $AB$ is opened.

The first line is the law of total probability.

$$P(Q)=P(Q|R)P(R)+P(Q|R')P(R')$$

and we have $P(R)=1-p$.

Hence we want to prove that $P(Q|R)=(1-p^2)^2$ and $P(Q|R')=1-(1-(1-p)^2)^2$.

If the event $R$ happens, that is if $BC$ is open, in order to reach $D$ from $A$. We need ($AB$ or $AC$ to be opened) and ($BD$ or $CD$ to be opened).

\begin{align} P(Q|R)&=P((O_{AB} \cup O_{AC}) \cap (O_{BD} \cup O_{CD})) \\ &=P(O_{AB} \cup O_{AC}) \cdot P(O_{BD} \cup O_{CD}) \\ &= \left( 1-P(O_{AB}')P(O_{AC}')\right)\left( 1-P(O_{BD}')P(O_{CD}')\right) \\ &= (1-p^2)^2 \end{align}

If the event $R'$ happens, that is if $BC$ is not open, in order to reach $D$ from $A$, we need ($AB$ and $BD$ to be opened) or ($AC$ and $CD$ to be opened).

\begin{align} P(Q|R') &= P((O_{AB} \cap O_{BD}) \cup (O_{AC} \cup O_{CD})) \\ &=1-P((O_{AB} \cap O_{BD})' )\cdot P( (O_{AC} \cup O_{CD})') \\ &= 1-(1-P(O_{AB})P(O_{BD}))\cdot (1-P(O_{AC})P(O_{CD}))\\ &= 1-(1-(1-p)^2)^2\\ \end{align}

• Thank you. For people who may need more detail for the second part of the solution, I would add the following: – HiMaths Sep 9 '18 at 8:03
• Following from the comment above I would add the following between the 1st and the 2nd lines of the second part of the solution: = 1 - P ( ( ( AB and BD ) or ( AC and CD ) ) ' ) and then = 1 - P ( ( AB and BD ) ' and ( AC and CD ) ' ) – HiMaths Sep 9 '18 at 8:12