91Ó°ÊÓ

The total number of domain names with just two letters (r=2) and characters (n=26) in a sequence may be computed as follows: number of potential outcomes = nr = 262 = 676 a2)

Assume that we can select first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways, and that for each selected element we can select the second element in \({{\rm{n}}_{\rm{2}}}\)ways, then we have the number of pairs to be \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

\({\rm{26*26 = 676}}\)

Domain names consisting of just two letters in sequence can be formed in ways, because \({{\rm{n}}_{\rm{1}}}{\rm{ = 26}}\)and \({{\rm{n}}_{\rm{2}}}{\rm{ = 26}}\)

If we allow digits as well, we'll get \({\rm{26 + 10 = 36}}\)characters, indicating that we can make a two-character sequence.

\({\rm{36*36 = 1296}}\) ways.

Domain names that consist of three letters in sequence may be made in the same way that domain names consisting of three letters in sequence can be formed in the same way that domain names consisting of three letters in sequence can be formed in the same way that domain names

\({\rm{26*26*26 = 17576}}\)

In the same way as in (a), when we add numbers, we get 36 characters.

\({\rm{36*36*36 = 46656}}\)

in a variety of ways,

To be more specific, for \({\rm{k = 3}}\), we utilize the following:

For k-Tuples, there is a Product Rule.

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\) ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

Similarly, there are \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}{\rm{* \ldots *}}{{\rm{n}}_{\rm{k}}}\) potential \({\rm{k}}\)-tuples given an ordered collection of \({\rm{k}}\) components, where the \({{\rm{k}}^{{\rm{th\;}}}}\) can be selected in \({{\rm{n}}_{\rm{k}}}\) ways.

Using the Product Rule for k-Tuplets, the number of ways to build a domain name with four letters in succession is \({\rm{k = }}\) \({\rm{4}}\)

\({\rm{26*26*26*26 = 456976}}\)

ways. In the same way, when we add digits, we get

\({\rm{36*36*36*36 = 1679616}}\)

The likelihood of a four-character name being held is

\(\begin{aligned}{{\rm{P(\{ four - character name is already owned\} )}}}&{\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1 - P(\{ four - character name is available\} )}}}\\{}&{\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{1 - }}\frac{{{\rm{97786}}}}{{{\rm{3}}{{\rm{6}}^{\rm{4}}}}}{\rm{ = 0}}{\rm{.942}}}\end{aligned}\)

(1): For each event A, $ \({\rm{P(A) + P}}\left( {{\rm{A'}}} \right){\rm{ = 1}}\), and the complement of the event that the four-character name is already taken is the event that the four-character name is taken;

(2): In the exercise, the number of good outcomes is provided, 97786, and all potential outcomes are stated in (c)

\({\rm{P(A) = }}\frac{{{\rm{\# favorable outcomes inA}}}}{{{\rm{\# outcomes in the sample space}}}}\)