# Schaum's Probability & Random Processes Question 1.11

• September 5th 2007, 09:09 AM
malweth
Schaum's Probability & Random Processes Question 1.11
I was going to post this yesterday, but thought I understood it ... and now I don't:

In Schaum's Probability, Random Variables, & Random Processes by Hwei Hsu, problem 1.11 gives four switched circuit diagrams, each with 3 switches. The goal is to define the path from $a$ to $b$. The text says:

Consider the switching networks shown in Fig. 1-5. Let $A_1$, $A_2$, and $A_3$ denote the events that the switches $s_1$, $s_2$, and $s_3$ are closed, respectively. Let $A_{ab}$ denote the event that there is a closed path between terminals $a$ and $b$. Express $A_{ab}$ in terms of $A_1$, $A_2$, and $A_3$ for each of the networks shown.

The two networks I'm concerned with are (a) and (b), which simply have the switches in series and parallel, respectively. The answers given are:

(a) $A_{ab} = A_1 \cap A_2 \cap A_3$
(b) $A_{ab} = A_1 \cup A_2 \cup A_3$

This doesn't make sense to me. It is true that in (a), all switches must be switched, and I would agree that $A_{ab} = \{A_1A_2A_3\}$ (in fact, that was my answer to the problem) but I don't see how $\{A_1A_2A_3\} = A_1 \cap A_2 \cap A_3$ unless $A_1 = A_2 = A_3$. Problem (b) has a similar issue.

Is the book wrong or do I not understand the intersection and union ( $\cap$/ $\cup$) correctly?
• September 5th 2007, 12:54 PM
topsquark
Quote:

Originally Posted by malweth
I was going to post this yesterday, but thought I understood it ... and now I don't:

In Schaum's Probability, Random Variables, & Random Processes by Hwei Hsu, problem 1.11 gives four switched circuit diagrams, each with 3 switches. The goal is to define the path from $a$ to $b$. The text says:

Consider the switching networks shown in Fig. 1-5. Let $A_1$, $A_2$, and $A_3$ denote the events that the switches $s_1$, $s_2$, and $s_3$ are closed, respectively. Let $A_{ab}$ denote the event that there is a closed path between terminals $a$ and $b$. Express $A_{ab}$ in terms of $A_1$, $A_2$, and $A_3$ for each of the networks shown.

The two networks I'm concerned with are (a) and (b), which simply have the switches in series and parallel, respectively. The answers given are:

(a) $A_{ab} = A_1 \cap A_2 \cap A_3$
(b) $A_{ab} = A_1 \cup A_2 \cup A_3$

This doesn't make sense to me. It is true that in (a), all switches must be switched, and I would agree that $A_{ab} = \{A_1A_2A_3\}$ (in fact, that was my answer to the problem) but I don't see how $\{A_1A_2A_3\} = A_1 \cap A_2 \cap A_3$ unless $A_1 = A_2 = A_3$. Problem (b) has a similar issue.

Is the book wrong or do I not understand the intersection and union ( $\cap$/ $\cup$) correctly?

Where's Figure 1-5!?

-Dan
• September 5th 2007, 06:05 PM
malweth
Quote:

The two networks I'm concerned with are (a) and (b), which simply have the switches in series and parallel, respectively.
I wouldn't know how to draw it.

Series means that you have:

$a$ -- $s_1$ -- $s_2$ -- $s_3$ -- $b$

(e.g. all three switches must be down to connect $a$ to $b$).

Parallel is harder to explain, but essentially:

$a$ --- $s_1$ --- $b$
$a$ --- $s_2$ --- $b$
$a$ --- $s_3$ --- $b$

(e.g. at least one of $s_1$, $s_2$, or $s_3$ must be down to complete the circuit).
• September 5th 2007, 07:08 PM
JakeD
Quote:

Originally Posted by malweth
I was going to post this yesterday, but thought I understood it ... and now I don't:

In Schaum's Probability, Random Variables, & Random Processes by Hwei Hsu, problem 1.11 gives four switched circuit diagrams, each with 3 switches. The goal is to define the path from $a$ to $b$. The text says:

Consider the switching networks shown in Fig. 1-5. Let $A_1$, $A_2$, and $A_3$ denote the events that the switches $s_1$, $s_2$, and $s_3$ are closed, respectively. Let $A_{ab}$ denote the event that there is a closed path between terminals $a$ and $b$. Express $A_{ab}$ in terms of $A_1$, $A_2$, and $A_3$ for each of the networks shown.

The two networks I'm concerned with are (a) and (b), which simply have the switches in series and parallel, respectively. The answers given are:

(a) $A_{ab} = A_1 \cap A_2 \cap A_3$
(b) $A_{ab} = A_1 \cup A_2 \cup A_3$

This doesn't make sense to me. It is true that in (a), all switches must be switched, and I would agree that $A_{ab} = \{A_1A_2A_3\}$ (in fact, that was my answer to the problem) but I don't see how $\{A_1A_2A_3\} = A_1 \cap A_2 \cap A_3$ unless $A_1 = A_2 = A_3$. Problem (b) has a similar issue.

Is the book wrong or do I not understand the intersection and union ( $\cap$/ $\cup$) correctly?

It seems to me you are not understanding the difference between events and sample points in a probability space.

A probability space is a triple $(\Omega,\mathcal{F},P).$ $\Omega$ is the space of sample points, here describing the state of the 3 switches. So I would imagine $\Omega$ consists of triples such as (1,0,1) which would say switches 1 and 3 are closed and switch 2 is open. $\mathcal{F}$ are the allowable subsets of $\Omega$, called the events. So the event that switches 1 and 3 are closed and switch 2 is open is the set $A = \{ (1,0,1) \}$. Notice that an event is a set of sample points.

The event that switch $s_i$ is closed would be $A_i = \{ (s_1,s_2,s_3) \in \Omega \ | \ s_i = 1 \ \}$. The event $A$ that switches 1 and 3 are closed and switch 2 is open can be written as $A = \{ (1,0,1) \} = A_1 \cap A_2^C \cap A_3.$ The event $A_{ab}$ switches 1, 2 and 3 are closed is $A_{ab} = \{ (1,1,1) \} = A_1 \cap A_2 \cap A_3$ as the book says. The event $A_{ab}$ that at least one of switches 1, 2 and 3 are closed is $A_{ab} = \{ (s_1,s_2,s_3) \in \Omega \ | \ s_1 = 1 \vee s_2 = 1 \vee s_3 = 1 \} = A_1 \cup A_2 \cup A_3$.