91影视

There are $n$ differing types of coupons,

Each with a distinct likelihood of being collected $鈭�p_i$

So every card is self-contained,

$n$cards get retrieved.

a)

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

$P\left(A_{蟽}\right)=p_1鈰�p_2鈰�鈥�鈰�p_n\mathrm{鈭赌}蟽鈭�S$

Here, the term "freedom" is applied.

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

$P\left(\underset{蟽鈭�S}{鈭�}鈥�A_{蟽}\right)=\underset{蟽鈭�S}{鈭�}鈥�p_1鈰�p_2鈰�鈥�鈰�p_n=n!\underset{i=1}{\overset{n}{鈭�}}鈥�p_i$

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

$p_1=p_2=鈥�=p_n=\frac{1}{n}$

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

Likely hood thatevery of the tickets retrieved wouldn't be of kind $i$if it's of form $iandj$:

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

We can even see this if the tickets are identity:

localid="1649661845457" $$\begin{gathered} \begin{array}{c}P\left(E_i\right)={\left(\frac{n鈭�1}{n}\right)}^n\\ P\left(E_i鈭�E_j\right)={\left(\frac{n鈭�2}{n}\right)}^n\\ P\left(E_i鈭�E_j鈭�E_k\right)={\left(\frac{n鈭�3}{n}\right)}^n\end{array} \\ .... \end{gathered}$$

$P\left(\underset{i=1}{\overset{n}{鈭�}}鈥�E_i\right)=\underset{k=1}{\overset{n}{鈭�}}鈥�\left[(鈭�1{)}^{k鈭�1}\underset{}{鈭�}鈥�P\left(E_{i_1}鈥�E_{i_k}\right)\right]$

$P\left(\underset{i=1}{\overset{n}{鈭�}}鈥�E_i\right)=\underset{k=1}{\overset{n}{鈭�}}鈥�\left[(鈭�1{)}^{k鈭�1}\underset{}{鈭�}鈥�{\left(\frac{n鈭�k}{n}\right)}^n\right]$

localid="1649662036901" $P\left(\underset{i=1}{\overset{n}{鈭�}}鈥�E_i\right)=\underset{k=1}{\overset{n}{鈭�}}鈥�\left[(鈭�1{)}^{k鈭�1}\left(\begin{array}{l}n\\ k\end{array}\right)脳{\left(\frac{n鈭�k}{n}\right)}^n\right]$

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

$P\left(\underset{i=1}{\overset{n}{鈭�}}鈥�E_i\right)=P\left[{\left(\underset{蟽鈭�S}{鈭�}鈥�A_{蟽}\right)}^c\right]=1鈭�P\left(\underset{蟽鈭�S}{鈭�}鈥�A_{蟽}\right)=1鈭�n!{\left(\frac{1}{n}\right)}^n$

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.

localid="1649729190091" $$\begin{gathered} \underset{k=1}{\overset{n}{鈭�}}鈥�\left[(鈭�1{)}^{k鈭�1}\left(\begin{array}{l}n\\ k\end{array}\right)脳{\left(\frac{n鈭�k}{n}\right)}^n\right] \\ =1鈭�n!{\left(\frac{1}{n}\right)}^n \end{gathered}$$

localid="1649729208706" $$\begin{gathered} \underset{k=1}{\overset{n}{鈭�}}鈥�\left[(鈭�1{)}^{k鈭�1}\left(\begin{array}{l}n\\ k\end{array}\right)脳(n鈭�k{)}^n\right] \\ =n^n鈭�n! \end{gathered}$$

localid="1649729225380" $n!=\underset{k=1}{\overset{n}{鈭�}}鈥�\left[(鈭�1{)}^{k鈭�1}\left(\begin{array}{l}n\\ k\end{array}\right)脳(n鈭�k{)}^n\right]鈭�n^n$

localid="1649729246136" $鈭�n^n=(鈭�1{)}^{0鈭�1}\left(\begin{array}{c}n\\ 0\end{array}\right)脳(n鈭�0{)}^n$

localid="1649729261777" $n!=\underset{k=0}{\overset{n}{鈭�}}鈥�\left[(鈭�1{)}^{k鈭�1}\left(\begin{array}{l}n\\ k\end{array}\right)脳(n鈭�k{)}^n\right]$

Iswhat's the specified individuality.