Question 1130394
<pre>
Imagine that all and each line are marked by the first 13 letters of English alphabet:

1  2  3  4  5  6  7  8  9  10  11  12  13
A  B  C  D  E  F  G  H  I   J   K   L   M 


Then the space of all possible events is the set of all 3-letter words comprising of these letters.

Repetitions of letters in these words are allowed.


It is easy to calculate the number of all such 3-letter words.

Any of 13 letter can stay in the 1-st position. This gives 13 opportunities.
Any of 13 letter can stay in the 2-nd position. This gives 13 opportunities.
Any of 13 letter can stay in the 3-rd position. This gives 13 opportunities.


In all, there are {{{13^3}}} such words.
Correspondingly, there are {{{13^3}}} elements in the space of events, in all.


Now, the favorable events are those 3-letter words what have no repetitions.

The number of such words is exactly  13*12*11 = 1716.



Therefore, the probability under the question is equal to  

{{{(13*12*11)/13^3}}} = 0.7811 = 78.11% (approximately).     <U>ANSWER</U>
</pre>

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Very similar to the problem 4 of the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Selected-probability-problems-from-the-archive.lesson>Selected probability problems from the archive</A> 

in this site.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic &nbsp;"<U>Solved problems on Probability</U>". 



Save the link to this textbook together with its description


Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson


into your archive and use when it is needed.