Model the situation with a Laplace room

[mathmari]

Hey!

An ice cream parlour offers 12 different types of ice cream, including vanilla ice cream. There are 8 people passing by, each of whom chosses a ball of ice cream. Of course, the ice cream parlour has taken good precautions, so that there is enough ice cream from each variety.
Model the situation with a Laplace room. To do this, enter an appropriate result set and the events of interest in (a) and (b) as a subset of the result set. Calculate the probability that
(a) each of 8 people order a different type of ice cream,
(b) exactly two of the 8 people would like to make vanilla ice cream.

I have done the following :

Let $\Omega := \{1, 2, \ldots , 12\}^8$, where $(\omega_1, \ldots , \omega_8) \in \Omega$ means that the $i$-th person chooses the flavour $\omega_i$ ($i \in \{1, \ldots , 8\}$).
Let $p(\omega) :=\frac{1}{|\Omega|}=\frac{1}{12^8}$ for all $\omega \in \Omega$.
Then $(\Omega, p)$ is the Laplace experiment.

Is that correctand compelte so far? :unsure: (a) As a subset of $\Omega$ we can write the event $A$ as follows :
\begin{equation*}A=\left \{(\omega_1, \ldots , \omega_8) \in \Omega \mid \omega_i\neq \omega_j \ \forall i,j \in \{1, \ldots , 8\} \text{ with } i\neq j \right \}\end{equation*}
Since $(\Omega, p)$ is a Laplace experiment, we have that\begin{equation*}P(A)=\frac{|A|}{|\Omega|}=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5}{12^8}\approx 0.046\end{equation*}
Is that correct? :unsure: (b) As a subset of $\Omega$ we can write the event $B$ as follows :
\begin{equation*}B=\left \{(\omega_1, \ldots , \omega_8) \in \Omega \mid \exists i,j,z\in \{1, \ldots 8\} : \omega_i=\omega_j="\text{vanilla}" \land \omega_z\neq "\text{vanilla}" \right \}\end{equation*} I am not really sure if we write this event in this way. :unsure:
Since $(\Omega, p)$ is a Laplace experiment, we have that\begin{equation*}P(B)=\frac{|B|}{|\Omega|}=\end{equation*} If the above is correct... How can we calculate $|B|$ ? Could you give me a hint? :unsure:

 

[I like Serena]

Model the situation with a Laplace room. To do this, enter an appropriate result set and the events of interest in (a) and (b) as a subset of the result set.

I have done the following :
Let $\Omega := \{1, 2, \ldots , 12\}^8$, where $(\omega_1, \ldots , \omega_8) \in \Omega$ means that the $i$-th person chooses the flavour $\omega_i$ ($i \in \{1, \ldots , 8\}$).
Let $p(\omega) :=\frac{1}{|\Omega|}=\frac{1}{12^8}$ for all $\omega \in \Omega$.
Then $(\Omega, p)$ is the Laplace experiment.

Hey mathmari!

I can't seem to find a definition of a "Laplace room" or "Laplace experiment".
Can you perhaps provide a reference or quote?

Either way, it looks to me that you defined the probability space correctly.

(a) As a subset of $\Omega$ we can write the event $A$ as follows :
\begin{equation*}A=\left \{(\omega_1, \ldots , \omega_8) \in \Omega \mid \omega_i\neq \omega_j \ \forall i,j \in \{1, \ldots , 8\} \text{ with } i\neq j \right \}\end{equation*}
Since $(\Omega, p)$ is a Laplace experiment, we have that\begin{equation*}P(A)=\frac{|A|}{|\Omega|}=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5}{12^8}\approx 0.046\end{equation*}

Yep. (Nod)

(b) As a subset of $\Omega$ we can write the event $B$ as follows :

How can we calculate $|B|$ ? Could you give me a hint?

What is the number of outcomes where customer 1 and 2 want vanilla and that none of the others want vanilla?
How many ways can we pick 2 customers from 8?

 

[mathmari]

I can't seem to find a definition of a "Laplace room" or "Laplace experiment".
Can you perhaps provide a reference or quote?

Either way, it looks to me that you defined the probability space correctly.

Laplace Room is a discrete probability space $(\Omega, P)$ with finite result set $\Omega$ and the discrete equidistance as probability measure $P$.
The Laplace space provides the mathematical model for describing a Laplace experiment, i.e. an experiment with a finite number of equally likely outcomes. Typical examples of Laplace experiments are dice with a fair cube or tossing an untvanked coin. Furthermore, Laplace rooms are usually used as a model in the drawing of so-called representative samples.

What is the number of outcomes where customer 1 and 2 want vanilla and that none of the others want vanilla?
How many ways can we pick 2 customers from 8?

Is it equal to $\binom{8}{2}$ ? :unsure:

 

[I like Serena]

Laplace Room is a discrete probability space $(\Omega, P)$ with finite result set $\Omega$ and the discrete equidistance as probability measure $P$.
The Laplace space provides the mathematical model for describing a Laplace experiment, i.e. an experiment with a finite number of equally likely outcomes. Typical examples of Laplace experiments are dice with a fair cube or tossing an untvanked coin. Furthermore, Laplace rooms are usually used as a model in the drawing of so-called representative samples.

All correct then. (Nod)

Is it equal to $\binom{8}{2}$ ?

That is indeed the number of ways we can pick the 2 customers that picked vanilla.
However, there are still many different outcomes depending on what the other customers pick. How many are those given the 2 customers that pick vanilla?

 

[mathmari]

All correct then. (Nod)

Ah is the set $B$ correct in the way I wrote it? :unsure:

That is indeed the number of ways we can pick the 2 customers that picked vanilla.
However, there are still many different outcomes depending on what the other customers pick. How many are those given the 2 customers that pick vanilla?

Is it maybe : $|B|=\binom{8}{2}\cdot \binom{12-1}{8-2}$ ? :unsure:

 

[I like Serena]

Ah is the set $B$ correct in the way I wrote it?

Ah no. It allows for a third customer to have vanilla, which should not be allowed. (Worried)

Is it maybe : $|B|=\binom{8}{2}\cdot \binom{12-1}{8-2}$ ?

Nope. (Shake)

Each of the other 6 customers has 12-1 choices and the ordering matters.
If two customers have different ice creams, we get a different outcome if they swap.
More generally, the binomial $\binom n k$ counts how many unordered outcomes there are, and it is also about drawing without replacement.
We have a different formula for ordered outcomes, and also if it is with replacement.
That is $\binom{12-1}{8-2}$ is the number of way that the remaining 6 customers pick different ice creams that are not vanilla. But that is not what we need.

 

[mathmari]

Ah no. It allows for a third customer to have vanilla, which should not be allowed. (Worried)

Ah so the part with $\omega_z$ is wrong, isn't it? :unsure:

Nope. (Shake)

Each of the other 6 customers has 12-1 choices and the ordering matters.
If two customers have different ice creams, we get a different outcome if they swap.
More generally, the binomial $\binom n k$ counts how many unordered outcomes there are, and it is also about drawing without replacement.
We have a different formula for ordered outcomes, and also if it is with replacement.
That is $\binom{12-1}{8-2}$ is the number of way that the remaining 6 customers pick different ice creams that are not vanilla. But that is not what we need.

So for ordered outcomes and with replacement do we use the formula $n^k$, i.e. $6^{11}$ ? :unsure:

 

[I like Serena]

Ah so the part with $\omega_z$ is wrong, isn't it?

Indeed.

So for ordered outcomes and with replacement do we use the formula $n^k$, i.e. $6^{11}$ ?

Yes.
But what you have there is the number of ways 11 customers can choose 6 icecreams. (Sweating)

 

[mathmari]

Indeed.

Should it maybe be :
\begin{equation*}B=\left \{(\omega_1, \ldots , \omega_8) \in \Omega \mid \exists i,j\in \{1, \ldots 8\} : \omega_i=\omega_j="\text{vanilla}" \land \forall z\neq \{i,j\} : \omega_z\neq "\text{vanilla}" \right \}\end{equation*}
Or is my approach completely wrong? :unsure:

Yes.
But what you have there is the number of ways 11 customers can choose 6 icecreams. (Sweating)

Ah yes, it is the other way around. It should be $11^6$. Each of the six people have $11$ options, right?

So is the total number of $B$ equal to $\binom{8}{2}\cdot 11^6$ ? :unsure:

 

[I like Serena]

Should it maybe be :
\begin{equation*}B=\left \{(\omega_1, \ldots , \omega_8) \in \Omega \mid \exists i,j\in \{1, \ldots 8\} : \omega_i=\omega_j="\text{vanilla}" \land \forall z\neq \{i,j\} : \omega_z\neq "\text{vanilla}" \right \}\end{equation*}
Or is my approach completely wrong?

Ah yes, it is the other way around. It should be $11^6$. Each of the six people have $11$ options, right?

So is the total number of $B$ equal to $\binom{8}{2}\cdot 11^6$ ?

All correct now. (Sun)

 

[mathmari]

All correct now. (Sun)

Therefore we get $$P(B)=\frac{\binom{8}{2}\cdot 11^6}{12^8}$$ Is that correct? :unsure:

 

[I like Serena]

Therefore we get $$P(B)=\frac{\binom{8}{2}\cdot 11^6}{12^8}$$ Is that correct?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Great! Thank you very much! (Sun)

 

[mathmari]

If two customers have different ice creams, we get a different outcome if they swap.

Why is the ordering important in this case? Do we not want just that all these $6$ people chooses something else than vanilla? :unsure:

 

[I like Serena]

Why is the ordering important in this case? Do we not want just that all these $6$ people chooses something else than vanilla?

Yes, but we are dividing by the total number of possible ordered ice cream distributions, which is what our sample space consists of.
It means we need to use the number of possible ordered ice cream distribution that match the event $B$.

 

[mathmari]

Yes, but we are dividing by the total number of possible ordered ice cream distributions, which is what our sample space consists of.
It means we need to use the number of possible ordered ice cream distribution that match the event $B$.

Could you explain that further to me? I haven't really understood that. :unsure:

 

[I like Serena]

Could you explain that further to me? I haven't really understood that.

You had for $B$:

Should it maybe be :
\begin{equation*}B=\left \{(\omega_1, \ldots , \omega_8) \in \Omega \mid \exists i,j\in \{1, \ldots 8\} : \omega_i=\omega_j="\text{vanilla}" \land \forall z\neq \{i,j\} : \omega_z\neq "\text{vanilla}" \right \}\end{equation*}

Each element of $B$ is of the form $(\omega_1, \ldots , \omega_8)$.
That is an ordered set aka tuple isn't it?

 

[mathmari]

Do we have that $\Omega := \{1, 2, \ldots , 12\}^8$ because the elements of $\Omega$ are the 8-tupels where each element can take a value between $1$ and $12$ ? :unsure:

 

[I like Serena]

Do we have that $\Omega := \{1, 2, \ldots , 12\}^8$ because the elements of $\Omega$ are the 8-tupels where each element can take a value between $1$ and $12$ ?

Yes. (Nod)