Expected number of sold books and profit

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In summary: Thinking) nevermind. We should start from 0 for (c). 🤔In summary, this conversation discusses a contract between a bookstore and a publisher regarding the sale of a book. The bookstore buys 15 copies of the book at 20 euros each and sells them for 30 euros each. The contract allows the bookstore to return unsold copies after a year for a refund of 18 euros each. The probability of selling a certain number of books is determined by a random variable X with a probability function of $\frac{x+2}{150}$ for $x \in \{1,2,...,15\}$. The conversation also includes discussions on verifying the probability function and determining the expected number of sold copies and
  • #1
mathmari
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Hey! :giggle:

A bookstore buys 15 copies of a book at a price of 20 euros each and offers them on sale for 30 euros each. The contract provides that the bookstore after a year can return the unsold copies of the book and receive 18 euros each.
Let the number of sold copies of this book in a year be determined by a random variable X with probability function $p_X(x)=\frac{x+2}{150}, \ x\in \{1,2, \ldots , 15\}$.
(a) Verify that $p_X$ is indeed a probability function.
(b) Determine the expected number of sold copies in a year. Write also the intermediate steps.
(c) Determine the bookstore's expected profit from the sale of this book if all unsold books are returned after one year. Write also the intermediate steps. I have done the following :
(a) Do we have to check that $p_X(\Omega)=1$ and $p_X\left (\cup A_i\right )=\sum p_X(A_i)$ ? :unsure:

(b) Is the expected number of sold copies equal to $E[X]=\sum_{i=1}^{15}x\cdot p_X(x)$ ? :unsure:

(c) The total expected profit is the sum of expected profits from each book, right? For one book teh expected profit is $(30-20)\cdot 0.50+(18-20)\cdot 0.50=4$, or isn't the probability to sell or not to sell equal to $\frac{1}{2}$ ? Is then the total profit $15\cdot 4=60$ ? :unsure:
 
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  • #2
mathmari said:
I have done the following :
(a) Do we have to check that $p_X(\Omega)=1$ and $p_X\left (\cup A_i\right )=\sum p_X(A_i)$ ?

(b) Is the expected number of sold copies equal to $E[X]=\sum_{i=1}^{15}x\cdot p_X(x)$ ?

Hey mathmari!

Yep. (Nod)

mathmari said:
(c) The total expected profit is the sum of expected profits from each book, right? For one book teh expected profit is $(30-20)\cdot 0.50+(18-20)\cdot 0.50=4$, or isn't the probability to sell or not to sell equal to $\frac{1}{2}$ ? Is then the total profit $15\cdot 4=60$ ?

Where did you get $0.50$ from? (Wondering)

The expected profit is $\sum \operatorname{profit}(x\text{ books sold}) \cdot p_X(x)$.
If we sold $x$ books then we gained $x\cdot 30 + (15-x)\cdot 18$ and we paid $30\cdot 20$.
So $\operatorname{profit}(x) =x\cdot 30 + (15-x)\cdot 18 - 30\cdot 20$. 🤔
 
  • #3
mathmari said:
(a) Do we have to check that $p_X(\Omega)=1$ and $p_X\left (\cup A_i\right )=\sum p_X(A_i)$ ? :unsure:

I got stuck right now. To show the first euality do we not use the second one? :unsure:
mathmari said:
(b) Is the expected number of sold copies equal to $E[X]=\sum_{i=1}^{15}x\cdot p_X(x)$ ? :unsure:

We have that $$E[X]=\sum_{x=1}^{15}x\cdot p_X(x)=\sum_{x=1}^{15}x\cdot \frac{x+2}{150}=\sum_{x=1}^{15}\frac{x^2+2x}{150}=\frac{148}{15}\approx 9,87$$ Is that correct? :unsure:
Klaas van Aarsen said:
The expected profit is $\sum \operatorname{profit}(x\text{ books sold}) \cdot p_X(x)$.
If we sold $x$ books then we gained $x\cdot 30 + (15-x)\cdot 18$ and we paid $30\cdot 20$.
So $\operatorname{profit}(x) =x\cdot 30 + (15-x)\cdot 18 - 30\cdot 20$. 🤔

Ah ok. So is the expected profit equal to $$\sum_{x=0}^{15}\left (x\cdot 30 + (15-x)\cdot 18 - 30\cdot 20\right )\cdot \frac{x+2}{150}=-216$$ ? In this case we start the sum from $0$, right? :unsure:
 
  • #4
mathmari said:
I got stuck right now. To show the first euality do we not use the second one?

Yes. We should list each possible outcome and verify that their probabilities sum to $1$. 🤔
mathmari said:
We have that $$E[X]=\sum_{x=1}^{15}x\cdot p_X(x)=\sum_{x=1}^{15}x\cdot \frac{x+2}{150}=\sum_{x=1}^{15}\frac{x^2+2x}{150}=\frac{148}{15}\approx 9,87$$ Is that correct?

Yep. (Nod)

mathmari said:
Ah ok. So is the expected profit equal to $$\sum_{x=0}^{15}\left (x\cdot 30 + (15-x)\cdot 18 - 30\cdot 20\right )\cdot \frac{x+2}{150}=-216$$ ? In this case we start the sum from $0$, right?

I made a mistake. We paid $15\cdot 20$ instead. (Blush)

And yes, we start the sum from $0$. We should also do that at (b), although the term with $x=0$ cancels anyway. 🤔
 
  • #5
Klaas van Aarsen said:
Yes. We should list each possible outcome and verify that their probabilities sum to $1$. 🤔

We have that \begin{align*}&p_X(1)=\frac{1+2}{150}=\frac{1}{50} \\ &p_X(2)=\frac{2+2}{150}=\frac{2}{75}\\ &p_X(3)=\frac{3+2}{150}=\frac{1}{30}\\ &p_X(4)=\frac{4+2}{150}=\frac{1}{25}\\ &p_X(5)=\frac{5+2}{150}=\frac{7}{150}\\ &p_X(6)=\frac{6+2}{150}=\frac{4}{75}\\ &p_X(7)=\frac{7+2}{150}=\frac{3}{50}\\ &p_X(8)=\frac{8+2}{150}=\frac{1}{15}\\ &p_X(9)=\frac{9+2}{150}=\frac{11}{150}\\ &p_X(10)=\frac{10+2}{150}=\frac{2}{25}\\ &p_X(11)=\frac{11+2}{150}=\frac{13}{150}\\ &p_X(12)=\frac{12+2}{150}=\frac{7}{75}\\ &p_X(13)=\frac{13+2}{150}=\frac{1}{10}\\ &p_X(14)=\frac{14+2}{150}=\frac{8}{75}\\ &p_X(15)=\frac{15+2}{150}=\frac{17}{150}\end{align*}
Summing these we get \begin{equation*}\frac{1}{50} +\frac{2}{75}+\frac{1}{30}+\frac{1}{25}+\frac{7}{150}+\frac{4}{75}+\frac{3}{50}+\frac{1}{15}+\frac{11}{150}+\frac{2}{25}+\frac{13}{150}+\frac{7}{75}+\frac{1}{10}+\frac{8}{75}+\frac{17}{150}=1\end{equation*} Does this follow then that $p_X$ is a probability function ? :unsure:

Klaas van Aarsen said:
And yes, we start the sum from $0$. We should also do that at (b), although the term with $x=0$ cancels anyway. 🤔

Ah can we start from $0$ although the $p_X(x)$ is defined for $x\in \{1, \ldots , 15\}$ ?
 
  • #6
mathmari said:
We have that \begin{align*}&p_X(1)=\frac{1+2}{150}=\frac{1}{50} \\ &p_X(2)=\frac{2+2}{150}=\frac{2}{75}\\ &p_X(3)=\frac{3+2}{150}=\frac{1}{30}\\ &p_X(4)=\frac{4+2}{150}=\frac{1}{25}\\ &p_X(5)=\frac{5+2}{150}=\frac{7}{150}\\ &p_X(6)=\frac{6+2}{150}=\frac{4}{75}\\ &p_X(7)=\frac{7+2}{150}=\frac{3}{50}\\ &p_X(8)=\frac{8+2}{150}=\frac{1}{15}\\ &p_X(9)=\frac{9+2}{150}=\frac{11}{150}\\ &p_X(10)=\frac{10+2}{150}=\frac{2}{25}\\ &p_X(11)=\frac{11+2}{150}=\frac{13}{150}\\ &p_X(12)=\frac{12+2}{150}=\frac{7}{75}\\ &p_X(13)=\frac{13+2}{150}=\frac{1}{10}\\ &p_X(14)=\frac{14+2}{150}=\frac{8}{75}\\ &p_X(15)=\frac{15+2}{150}=\frac{17}{150}\end{align*}
Summing these we get \begin{equation*}\frac{1}{50} +\frac{2}{75}+\frac{1}{30}+\frac{1}{25}+\frac{7}{150}+\frac{4}{75}+\frac{3}{50}+\frac{1}{15}+\frac{11}{150}+\frac{2}{25}+\frac{13}{150}+\frac{7}{75}+\frac{1}{10}+\frac{8}{75}+\frac{17}{150}=1\end{equation*} Does this follow then that $p_X$ is a probability function ?

From the definition:
The probability measure $P:\mathcal{F}\to[0,1]$ is a function on $\mathcal{F}$ such that:
* $P$ is countably additive (also called σ-additive): if $\{A_i\}_{i=1}^\infty\subseteq\mathcal{F}$ is a countable collection of pairwise disjoint sets, then $\textstyle P(\bigcup_{i=1}^\infty A_i)=\sum_{i=1}^\infty P(A_i),$
* The measure of entire sample space is equal to one: $P(\Omega)=1$.

It should suffice that the sum of the probabilities of all outcomes (that are disjoint), sums up to 1 as it does. 🤔

mathmari said:
Ah can we start from $0$ although the $p_X(x)$ is defined for $x\in \{1, \ldots , 15\}$ ?
Hmm... actually we can't. (Shake)
That definition implies that we sell at least 1 book.
You verified that the sum of all probabilities starting from 1 book sold, does sum up to 1.
In other words, it is indeed not possible to sell 0 books - or rather, the probability on it is 0, which is kind of surprising. 🤔
It also means that for (c) we should start from 1.
 

What is the expected number of books that will be sold?

The expected number of books that will be sold is determined by various factors such as market demand, marketing strategies, and competition. A thorough market analysis and forecasting can help estimate the expected number of book sales.

What is the expected profit from selling the books?

The expected profit from selling the books is also influenced by several factors, including the cost of production, marketing expenses, and the selling price of the book. A detailed financial analysis can help determine the expected profit from selling the books.

How accurate are the expectations for book sales and profit?

The accuracy of the expectations for book sales and profit depends on the quality of the market analysis and forecasting. It is essential to gather reliable data and use appropriate methods to make accurate predictions.

What can be done to increase the expected number of book sales?

To increase the expected number of book sales, various strategies can be implemented, such as targeted marketing campaigns, collaborations with influencers or other authors, and offering promotions or discounts. It is also crucial to continuously monitor and adapt to market trends.

What are the potential risks that could affect the expected number of book sales and profit?

Some potential risks that could affect the expected number of book sales and profit include changes in market demand, unexpected competition, and production or distribution issues. It is important to have contingency plans in place to mitigate these risks and minimize their impact.

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