Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 29E from Chapter 2 from Devore's Probability and Statistics for Engineering and Science.
Problem 29E
Chapter:
Problem:
As of April 2006, roughly 50 million .com web doma...
Step-by-Step Solution
Given Information
As of April 2006, roughly 50 million .com web domain names were registered
Step-1:
Domain names consisting of just two letters in sequence
There are 26 characters in English. For the first place we can have any of the 26 characters and similarly for the second place also, we can have one of the 26 characters
Total number of domains with two characters
\[\begin{aligned}n &=26^{2} \\&=26 \times 26 \\&=676\end{aligned}\]When digits are also allowed, there are (26+10 = 36)
\[\begin{aligned}n &=36^{2} \\&=36 \times 36 \\&=1,296\end{aligned}\]
Step-2: Part (b)
Domain names consisting of just three letters in sequence
There are 26 characters in English. For the first place we can have any of the 26 characters and similarly for the second place also, we can have one of the 26 characters. Similarly for the third case also
Total number of domains with three characters
\begin{aligned}n &=26^{3} \\&=26 \times 26 \times 26 \\&=17,576\end{aligned}When digits are also allowed, there are (26+10 = 36)
\[\begin{aligned}n &=36^{3} \\&=36 \times 36 \times 36 \\&=46,656\end{aligned}\]
Step-3: Part (c)
Domain names consisting of just Four letters in sequence
Similar to the previous case \\Total number of domains with Four characters
\begin{aligned}n &=26^{4} \\&=26 \times 26 \times 26 \times 26 \\&= 456976\end{aligned}When digits are also allowed, there are (26+10 = 36)
\[\begin{aligned}n &=36^{4} \\&=36 \times 36 \times 36 \times 36 \\&=1679616 \end{aligned}\]
Step-4: Part (d)
Given that 97,786 of the four-character sequences using either letters or digits had not yet been claimed.
The digits already claimed are
\[\begin{array}{l}N = 1679616 - 97786\\\\N = 1581830\end{array}\]The probability that a randomly selected four-character sequences using either letters or digits is already owned is,
\[\begin{aligned}P(\text { digits already owned }) &=\dfrac{15,81,830}{1,679,616} \\&=0.941781 \\& \approx 0.942\end{aligned}\]==============================================================================================================
As of April 2006, roughly 50 million .com web domain names were registered
Step-1:
Domain names consisting of just two letters in sequence
There are 26 characters in English. For the first place we can have any of the 26 characters and similarly for the second place also, we can have one of the 26 characters
Total number of domains with two characters
\[\begin{aligned}n &=26^{2} \\&=26 \times 26 \\&=676\end{aligned}\]When digits are also allowed, there are (26+10 = 36)
\[\begin{aligned}n &=36^{2} \\&=36 \times 36 \\&=1,296\end{aligned}\]
Step-2: Part (b)
Domain names consisting of just three letters in sequence
There are 26 characters in English. For the first place we can have any of the 26 characters and similarly for the second place also, we can have one of the 26 characters. Similarly for the third case also
Total number of domains with three characters
\begin{aligned}n &=26^{3} \\&=26 \times 26 \times 26 \\&=17,576\end{aligned}When digits are also allowed, there are (26+10 = 36)
\[\begin{aligned}n &=36^{3} \\&=36 \times 36 \times 36 \\&=46,656\end{aligned}\]
Step-3: Part (c)
Domain names consisting of just Four letters in sequence
Similar to the previous case \\Total number of domains with Four characters
\begin{aligned}n &=26^{4} \\&=26 \times 26 \times 26 \times 26 \\&= 456976\end{aligned}When digits are also allowed, there are (26+10 = 36)
\[\begin{aligned}n &=36^{4} \\&=36 \times 36 \times 36 \times 36 \\&=1679616 \end{aligned}\]
Step-4: Part (d)
Given that 97,786 of the four-character sequences using either letters or digits had not yet been claimed.
The digits already claimed are
\[\begin{array}{l}N = 1679616 - 97786\\\\N = 1581830\end{array}\]The probability that a randomly selected four-character sequences using either letters or digits is already owned is,
\[\begin{aligned}P(\text { digits already owned }) &=\dfrac{15,81,830}{1,679,616} \\&=0.941781 \\& \approx 0.942\end{aligned}\]==============================================================================================================