Question 1175160
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            The solution by @math_tutor is perfectly correct.


            I want to show you another way of thinking.



<pre>
Having 5 different digits, you can form  5! = 1*2*3*4*5 = 120 different numbers.


In this soup of numbers, the last digit can be any of the given 5 digits with the same probability (obviously !).


Therefore, the amount of such numbers with any fixed last digit is  {{{1/5}}}  of the total amount of the numbers.


It means that the probability under the problem question is  {{{1/5}}}.        <U>ANSWER</U>
</pre>

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Thinking this way, you may solve other similar problems VERY QUICKLY without making unnecessary calculations.


But you should be aware about APPLICABILITY of this way thinking (!) - it depends on the problem (!)