Expected number of sold books and profit

[mathmari]

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 the 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:

 

[I like Serena]

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)

(c) The total expected profit is the sum of expected profits from each book, right? For one book the 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 $\operatorname{profit}(x) =x\cdot 30 + (15-x)\cdot 18 - 30\cdot 20$.

 

[mathmari]

(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:

(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:

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:

 

[I like Serena]

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 $1$.

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)

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.

 

[mathmari]

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:

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 $x\in \{1, \ldots , 15\}$ ?

 

[I like Serena]

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.

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.