Representing Faults in Manufacturing Departments Using Events

[mathmari]

Hey!

I am looking the following exercise:

A medium-sized company has $n = 3$ manufacturing departments. Faults in the production process can occur in these departments.
We have the following events:
\begin{align*}&A=\{"\text{All departments work without faults}"\} \\ &B=\{"\text{
no department works without faults}"\} \\ &C=\{"\text{
at least one department works without faults}"\} \\ &D=\{"\text{at most one department works without faults}"\} \\ &E=\{"\text{exactly one department works without faults}"\}\end{align*}

We also have the events $$F_k=\{"\text{the } k\text{-th department works without faults}"\}, \ \ k=1, 2, \ldots , n$$

I want to represent the events $A, B, \ldots , E$ using the events $F_k$ and simplify them. I have done the following:

\begin{align*}A=&\{"\text{All departments work without faults}"\} \\ =& \{"\text{the first department works without faults AND the second department works without faults } \\ & \text{AND the third department works without faults}"\} \\ =& \{"\text{the first department works without faults}"\} \cap\{\text{"the second department works without faults}"\} \\ & \cap \{"\text{the third department works without faults}"\}\\ =& F_1\cap F_2\cap F_3\\ = &\bigcap_{k=1}^3F_k\end{align*}

\begin{align*}B=&\{"\text{
no department works without faults}"\}\\ =& \{"\text{the first department works with faults AND the second department works with faults } \\ & \text{AND the third department works with faults}"\} \\ =& \{"\text{the first department works with faults}"\} \cap\{\text{"the second department works with faults}"\} \\ & \cap \{"\text{the third department works with faults}"\} \\ =&
\overline{\{"\text{the first department works without faults}"\}} \cap\overline{\{\text{"the second department works without faults}"\}} \\ & \cap \overline{\{"\text{the third department works without faults}"\}} \\ =& \overline{F_1}\cap \overline{F_2}\cap \overline{F_3}\\ =& \bigcap_{k=1}^3\overline{F_k}\end{align*}

\begin{align*}C=&\{"\text{
at least one department works without faults}"\}\\ = & \{"\text{the first department works without faults OR the second department works without faults } \\ & \text{OR the third department works without faults}"\}\\ = & \{"\text{the first department works without faults}"\} \cup\{\text{"the second department works without faults}"\} \\ & \cup \{"\text{the third department works without faults}"\} \\ =& F_1\cup F_2\cup F_3\\ = & \bigcup_{k=1}^3F_k\end{align*}

\begin{align*}D=&\{"\text{at most one department works without faults}"\} \\ =&\{"\text{no department works without faults}"\}\cup \{"\text{exactly one department works without faults}"\}\\ = &B\cup E\end{align*}

\begin{align*}E=&\{"\text{exactly one department works without faults}"\}\\ =&\left (F_1\cap \overline{F_2}\cap \overline{F_3}\right )\cup \left (\overline{F_1}\cap F_2\cap \overline{F_3}\right ) \cup \left (\overline{F_1}\cap \overline{F_2}\cap F_3\right )\end{align*} Is everything correct?

I am not sure about $D$ because I use the event $E$ that I define at the next step.

(Wondering)

 

[I like Serena]

Is everything correct?

I am not sure about $D$ because I use the event $E$ that I define at the next step.

Hey mathmari! (Smile)

It all seems correct to me.
As for D, shouldn't we substitute both B and E and try to simplify? (Wondering)

 

[mathmari]

As for D, shouldn't we substitute both B and E and try to simplify? (Wondering)

We have that \begin{equation*}D= \left (\overline{F_1}\cap \overline{F_2}\cap \overline{F_3}\right )\cup \left (F_1\cap \overline{F_2}\cap \overline{F_3}\right )\cup \left (\overline{F_1}\cap F_2\cap \overline{F_3}\right ) \cup \left (\overline{F_1}\cap \overline{F_2}\cap F_3\right )\end{equation*}

Considering first the first two parentheses we have the following: \begin{align*} &\left (\overline{F_1}\cap \overline{F_2}\cap \overline{F_3}\right )\cup \left (F_1\cap \overline{F_2}\cap \overline{F_3}\right ) \\ = &\left [\left (\overline{F_1}\cap \overline{F_2}\cap \overline{F_3}\right )\cup F_1\right ] \cap \left [\left (\overline{F_1}\cap \overline{F_2}\cap \overline{F_3}\right ) \cup \overline{F_2} \right ]\cap \left [\left (\overline{F_1}\cap \overline{F_2}\cap \overline{F_3}\right )\cup \overline{F_3}\right ] \\ =& \left [\left (\overline{F_1}\cup F_1\right )\cap \left (\overline{F_2}\cup F_1\right )\cap \left (\overline{F_3}\cup F_1\right )\right ] \cap \left [\left (\overline{F_1} \cup \overline{F_2}\right )\cap \left (\overline{F_2} \cup \overline{F_2}\right )\cap \left (\overline{F_3} \cup \overline{F_2}\right ) \right ] \\ & \cap \left [\left (\overline{F_1}\cup \overline{F_3}\right )\cap \left (\overline{F_2}\cup \overline{F_3}\right )\cap \left (\overline{F_3}\cup \overline{F_3}\right )\right ] \\ = & \left [\Omega\cap \left (\overline{F_2}\cup F_1\right )\cap \left (\overline{F_3}\cup F_1\right )\right ] \cap \left [\left (\overline{F_1} \cup \overline{F_2}\right )\cap \Omega\cap \left (\overline{F_3} \cup \overline{F_2}\right ) \right ] \cap \left [\left (\overline{F_1}\cup \overline{F_3}\right )\cap \left (\overline{F_2}\cup \overline{F_3}\right )\cap \Omega\right ] \\ = & \left [\left (\overline{F_2}\cup F_1\right )\cap \left (\overline{F_3}\cup F_1\right )\right ] \cap \left [\left (\overline{F_1} \cup \overline{F_2}\right )\cap \left (\overline{F_3} \cup \overline{F_2}\right ) \right ] \cap \left [\left (\overline{F_1}\cup \overline{F_3}\right )\cap \left (\overline{F_2}\cup \overline{F_3}\right )\right ] \\ = & \left (\overline{F_2}\cup F_1\right )\cap \left (\overline{F_3}\cup F_1\right ) \cap \left (\overline{F_1} \cup \overline{F_2}\right )\cap \left (\overline{F_3} \cup \overline{F_2}\right ) \cap \left (\overline{F_1}\cup \overline{F_3}\right )\cap \left (\overline{F_2}\cup \overline{F_3}\right )\end{align*}

Is this correct so far? Can we simplify that part further? (Wondering)

 

[I like Serena]

Can't we make it:
$$
(\bar F_1 \cap \bar F_2) \cup (\bar F_1 \cap \bar F_3) \cup (\bar F_2 \cap \bar F_3) =
\bigcup_{i=1}^3 \bigcap_{j\ne i} \overline F_j
$$

That seems simpler doesn't it? (Wondering)

 

[mathmari]

Can't we make it:
$$
(\bar F_1 \cap \bar F_2) \cup (\bar F_1 \cap \bar F_3) \cup (\bar F_2 \cap \bar F_3) =
\bigcup_{i=1}^3 \bigcap_{j\ne i} \overline F_j
$$

How do we get that? I got stuck right now. (Wondering)

 

[I like Serena]

How do we get that? I got stuck right now. (Wondering)

To be honest, I used a Karnaugh Map to figure it out. (Sweating)
But we can also see it directly, since what we need is that at least n-1=2 departments have no faults.

 

[mathmari]

To be honest, I used a Karnaugh Map to figure it out. (Sweating)
But we can also see it directly, since what we need is that at least n-1=2 departments have no faults.

Ah we have that \begin{align*}D= &\{\text{at most one department works without faults}\}\\ = &\{\text{no department works without faults OR exactly one department works without faults}\} \\ = &\{\text{all three department work with faults OR two department work with faults}\} \\ = & \{\text{at least two department work with faults}\} \\ = & \{\text{first AND
second dep. work with faults OR first AND third dep. work with faults} \\ & \text{OR second AND third dep. work with faults}\} \\ = & \left (\overline{F_1}\cap \overline{F_2}\right )\cup \left (\overline{F_1}\cap \overline{F_3}\right )\cup \left (\overline{F_2}\cap \overline{F_3}\right )\end{align*} right? (Wondering)

$x\in \left (\overline{F_1}\cap \overline{F_2}\right )\cup \left (\overline{F_1}\cap \overline{F_3}\right )\cup \left (\overline{F_2}\cap \overline{F_3}\right )$ means that $x$ is contained in at least one parenthesis. If is contained in one parenthesis then we have that exactly two departments work with faults and if $x$ is contained in two or three parenthesis then we have that exactly three departments work with faults, which means that at least two departments work with faults.
Is this correct? (Wondering)

\begin{align*}E=&\{"\text{exactly one department works without faults}"\}\\ =&\left (F_1\cap \overline{F_2}\cap \overline{F_3}\right )\cup \left (\overline{F_1}\cap F_2\cap \overline{F_3}\right ) \cup \left (\overline{F_1}\cap \overline{F_2}\cap F_3\right )\end{align*}

Can we simplify this expression? (Wondering)

 

[I like Serena]

Is this correct?

Can we simplify this expression?

Looks all correct to me.
And no, I don't think we can simplify it further. (Mmm)

 

[mathmari]

Looks all correct to me.

Great! (Clapping)

And no, I don't think we can simplify it further. (Mmm)

Can we write it also in the following form?
\begin{align*}E=&\{"\text{exactly one department works without faults}"\}\\ =&\left (F_1\cap \overline{F_2}\cap \overline{F_3}\right )\cup \left (\overline{F_1}\cap F_2\cap \overline{F_3}\right ) \cup \left (\overline{F_1}\cap \overline{F_2}\cap F_3\right )\\ =&\bigcup_{i=1}^3 F_i\bigcap_{j\neq i} \overline{F_j} \end{align*}
Or is it better to leave it as it is in the second line? (Wondering)

 

[I like Serena]

Can we write it also in the following form?
\begin{align*}E=&\{"\text{exactly one department works without faults}"\}\\ =&\left (F_1\cap \overline{F_2}\cap \overline{F_3}\right )\cup \left (\overline{F_1}\cap F_2\cap \overline{F_3}\right ) \cup \left (\overline{F_1}\cap \overline{F_2}\cap F_3\right )\\ =&\bigcup_{i=1}^3 F_i\bigcap_{j\neq i} \overline{F_j} \end{align*}
Or is it better to leave it as it is in the second line? (Wondering)

Sure, although shouldn't it be:
$$\bigcup_{i=1}^3 \left( F_i \cap \left(\bigcap_{j\neq i} \overline{F_j}\right) \right)$$
(Wondering)

And it's up to you to specify it or not. (Wink)

 

[mathmari]

Sure, although shouldn't it be:
$$\bigcup_{i=1}^3 \left( F_i \cap \left(\bigcap_{j\neq i} \overline{F_j}\right) \right)$$
(Wondering)

Ah yes (Blush)

And it's up to you to specify it or not. (Wink)

Ok!

Thank you so much! (Yes)