Q. 3.29 - A First Course in Probability

Question

There are $n$ distinct types of coupons, and each coupon obtained is, independently of prior types collected, of type $i$ with probability $p_i,\sum _{i=1}^n p_i=1$.

(a) If $n$ coupons are collected, what is the probability that one of each type is obtained?

(b) Now suppose that $p_1=p_2=\cdots =p_n=1/n$. Let $E_i$ be the event that there are no type $i$ coupons among the $n$ collected. Apply the inclusion–exclusion identity for the probability of the union of events to $P\left({\cup }_i{E}_i\right)$ to prove the identity

$n!=\sum _{k=0}^n (-1{)}^k\left(\begin{array}{l}n\\ k\end{array}\right)(n-k{)}^n$

Step-by-Step Solution

Answer

Add up the the probabilities  of the rewards being gathered in a very precise order. Make independent use of several coupons you've gathered.

$P=n!\cdot \prod _{i=1}^n p_i$

 The subsequent is that the results of  the inclusion-exclusion formula:

$P\left(\underset{i=1}{\overset{n}{\cup }} E_i\right)=\sum _{k=1}^n \left[(-1{)}^{k-1}\left(\begin{array}{l}n\\ k\end{array}\right)\cdot {\left(\frac{n-k}{n}\right)}^n\right]$

This is now the converse of the occurrence in (a), and summing the $2$ possibilities (addition of (a) and just last equation) provides the intended identity.

Step 1: Given

There are $n$ differing types of coupons,

Each with a distinct likelihood of being collected $-p_i$

 So every card is self-contained, 

 $n$ cards get retrieved.

a) 

The likelihood that discounts get retrieved inside this fashion (format event $A_{\sigma }$) by each variation of numerals $\{1,2,\dots ,n\}$ (in nomenclature $\sigma \in S$) is:

$P\left(A_{\sigma }\right)=p_1\cdot p_2\cdot \dots \cdot p_n\mathrm{\forall }\sigma \in S$

Here, the term "freedom" is applied.

while there are $n!$ variants, the outcomes that belong against them are strictly distinctive:

$P\left(\underset{\sigma \in S}{\cup } A_{\sigma }\right)=\sum _{\sigma \in S} p_1\cdot p_2\cdot \dots \cdot p_n=n!\prod _{i=1}^n p_i$

This can be the specified  likelihood of getting all sorts of  tickets.

Step 2: Inclusion and Exclusion Formula

$p_1=p_2=\dots =p_n=\frac{1}{n}$

Kind $i$ credits weren't retrieved by $E_i$.

Likely hood that every of the tickets retrieved wouldn't be of kind $i$ if it's of form $i and j$:

$1-p_i=\frac{n-1}{n},1-\left(p_i+p_j\right)=\frac{n-2}{n}$

We can even see this if the tickets are identity:

$$\begin{gathered} \begin{array}{c}P\left(E_i\right)={\left(\frac{n-1}{n}\right)}^n\\ P\left(E_i\cap E_j\right)={\left(\frac{n-2}{n}\right)}^n\\ P\left(E_i\cap E_j\cap E_k\right)={\left(\frac{n-3}{n}\right)}^n\end{array} \\ .... \end{gathered}$$

for each activity, there'll  be an inclusion-exclusion equation,

$P\left(\underset{i=1}{\overset{n}{\cup }} E_i\right)=\sum _{k=1}^n \left[(-1{)}^{k-1}\sum _{} P\left(E_{i_1}\dots E_{i_k}\right)\right]$

$P\left(\underset{i=1}{\overset{n}{\cup }} E_i\right)=\sum _{k=1}^n \left[(-1{)}^{k-1}\sum _{} {\left(\frac{n-k}{n}\right)}^n\right]$

$P\left(\underset{i=1}{\overset{n}{\cup }} E_i\right)=\sum _{k=1}^n \left[(-1{)}^{k-1}\left(\begin{array}{l}n\\ k\end{array}\right)\times {\left(\frac{n-k}{n}\right)}^n\right]$

Step 3: Complement Event

This may be the likelihood that it's at minimum single reasonably   coupon will never be received, it is the counterpart of both the particular  event

$P\left(\underset{i=1}{\overset{n}{\cup }} E_i\right)=P\left[{\left(\underset{\sigma \in S}{\cup } A_{\sigma }\right)}^c\right]=1-P\left(\underset{\sigma \in S}{\cup } A_{\sigma }\right)=1-n!{\left(\frac{1}{n}\right)}^n$

Part ($a$) and specified $p1,p2,......$ are utilized in final parity.

We get identity by joining this calculation with the previous  expression within the earlier row.

$$\begin{gathered} \sum _{k=1}^n \left[(-1{)}^{k-1}\left(\begin{array}{l}n\\ k\end{array}\right)\times {\left(\frac{n-k}{n}\right)}^n\right] \\ =1-n!{\left(\frac{1}{n}\right)}^n \end{gathered}$$

$$\begin{gathered} \sum _{k=1}^n \left[(-1{)}^{k-1}\left(\begin{array}{l}n\\ k\end{array}\right)\times (n-k{)}^n\right] \\ =n^n-n! \end{gathered}$$

$n!=\sum _{k=1}^n \left[(-1{)}^{k-1}\left(\begin{array}{l}n\\ k\end{array}\right)\times (n-k{)}^n\right]-n^n$

$-n^n=(-1{)}^{0-1}\left(\begin{array}{c}n\\ 0\end{array}\right)\times (n-0{)}^n$

$n!=\sum _{k=0}^n \left[(-1{)}^{k-1}\left(\begin{array}{l}n\\ k\end{array}\right)\times (n-k{)}^n\right]$

Is what's the specified individuality.