4 letters and 4 numbers to choose from, whats the probability that a specific 4 letter word is chosen












2












$begingroup$


We have 4 letters I, D, L, E and 4 numbers 1, 9, 8, 7 to create an 8 character password. What is the probability that in a randomly created password, "IDLE" appears?
(Order matters. ex: idle is different from ldie)






probability







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  • 2




    $begingroup$
    Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    7 hours ago










  • $begingroup$
    Can a character be used more than once?
    $endgroup$
    – saulspatz
    7 hours ago










  • $begingroup$
    Does "ID1987LE" count or must the four characters be consecutive ?
    $endgroup$
    – WW1
    7 hours ago










  • $begingroup$
    Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
    $endgroup$
    – Sujit Bhattacharyya
    7 hours ago
















2












$begingroup$


We have 4 letters I, D, L, E and 4 numbers 1, 9, 8, 7 to create an 8 character password. What is the probability that in a randomly created password, "IDLE" appears?
(Order matters. ex: idle is different from ldie)






probability







share|cite|improve this question









New contributor




Priscilla96 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    7 hours ago










  • $begingroup$
    Can a character be used more than once?
    $endgroup$
    – saulspatz
    7 hours ago










  • $begingroup$
    Does "ID1987LE" count or must the four characters be consecutive ?
    $endgroup$
    – WW1
    7 hours ago










  • $begingroup$
    Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
    $endgroup$
    – Sujit Bhattacharyya
    7 hours ago














2












2








2





$begingroup$


We have 4 letters I, D, L, E and 4 numbers 1, 9, 8, 7 to create an 8 character password. What is the probability that in a randomly created password, "IDLE" appears?
(Order matters. ex: idle is different from ldie)






probability







share|cite|improve this question









New contributor




Priscilla96 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




We have 4 letters I, D, L, E and 4 numbers 1, 9, 8, 7 to create an 8 character password. What is the probability that in a randomly created password, "IDLE" appears?
(Order matters. ex: idle is different from ldie)






probability




probability






share|cite|improve this question









New contributor




Priscilla96 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Priscilla96 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









GNUSupporter 8964民主女神 地下教會

12.8k72445




12.8k72445






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asked 8 hours ago









Priscilla96Priscilla96

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  • 2




    $begingroup$
    Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    7 hours ago










  • $begingroup$
    Can a character be used more than once?
    $endgroup$
    – saulspatz
    7 hours ago










  • $begingroup$
    Does "ID1987LE" count or must the four characters be consecutive ?
    $endgroup$
    – WW1
    7 hours ago










  • $begingroup$
    Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
    $endgroup$
    – Sujit Bhattacharyya
    7 hours ago














  • 2




    $begingroup$
    Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts.
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    7 hours ago










  • $begingroup$
    Can a character be used more than once?
    $endgroup$
    – saulspatz
    7 hours ago










  • $begingroup$
    Does "ID1987LE" count or must the four characters be consecutive ?
    $endgroup$
    – WW1
    7 hours ago










  • $begingroup$
    Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
    $endgroup$
    – Sujit Bhattacharyya
    7 hours ago








2




2




$begingroup$
Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
7 hours ago




$begingroup$
Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
7 hours ago












$begingroup$
Can a character be used more than once?
$endgroup$
– saulspatz
7 hours ago




$begingroup$
Can a character be used more than once?
$endgroup$
– saulspatz
7 hours ago












$begingroup$
Does "ID1987LE" count or must the four characters be consecutive ?
$endgroup$
– WW1
7 hours ago




$begingroup$
Does "ID1987LE" count or must the four characters be consecutive ?
$endgroup$
– WW1
7 hours ago












$begingroup$
Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
$endgroup$
– Sujit Bhattacharyya
7 hours ago




$begingroup$
Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
$endgroup$
– Sujit Bhattacharyya
7 hours ago










2 Answers
2






active

oldest

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3












$begingroup$

The number of all possible passwords with these 8 characters is 8!.



The number of passwords where I,D,L,E appear consecutively is equal to the number of possible positions of IDLE within the password (which is 4) times the number of possible permutations of the remaining 4 characters (which is 4!).



So the probability will be $frac{4(4!)}{8!}$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    the number of possible positions of IDLE within the password should be 5.
    $endgroup$
    – Julius
    5 hours ago



















1












$begingroup$

Your question can be interpreted in two different ways. The way of choosing the password as a random permutation of "IDLE1987" was dealt in the answer by Zaeem Hussein. Here I deal with the other interpretation, where the letters of the password are chosen independently and uniformly from $8$ available ones.



Suppose, we generated a password and a subword "IDLE" appeared in it. If the password has length 8, then the letter 'I' of "IDLE" is either first in the passport, or second, or third, or fourth ($4$ different cases). In any of those cases only it and the $3$ letters after it matter, so the probability of each of these situations is $frac{1}{8^4}$. However, one does not simply sum them up in order to get the answer, as then we will count one possible password twice. That is "IDLEIDLE". Its probability is $frac{1}{8^8}$. So the final probability is $frac{4}{8^4} - frac{1}{8^8} = frac{16383}{16777216}$






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

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    3












    $begingroup$

    The number of all possible passwords with these 8 characters is 8!.



    The number of passwords where I,D,L,E appear consecutively is equal to the number of possible positions of IDLE within the password (which is 4) times the number of possible permutations of the remaining 4 characters (which is 4!).



    So the probability will be $frac{4(4!)}{8!}$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      the number of possible positions of IDLE within the password should be 5.
      $endgroup$
      – Julius
      5 hours ago
















    3












    $begingroup$

    The number of all possible passwords with these 8 characters is 8!.



    The number of passwords where I,D,L,E appear consecutively is equal to the number of possible positions of IDLE within the password (which is 4) times the number of possible permutations of the remaining 4 characters (which is 4!).



    So the probability will be $frac{4(4!)}{8!}$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      the number of possible positions of IDLE within the password should be 5.
      $endgroup$
      – Julius
      5 hours ago














    3












    3








    3





    $begingroup$

    The number of all possible passwords with these 8 characters is 8!.



    The number of passwords where I,D,L,E appear consecutively is equal to the number of possible positions of IDLE within the password (which is 4) times the number of possible permutations of the remaining 4 characters (which is 4!).



    So the probability will be $frac{4(4!)}{8!}$.






    share|cite|improve this answer









    $endgroup$



    The number of all possible passwords with these 8 characters is 8!.



    The number of passwords where I,D,L,E appear consecutively is equal to the number of possible positions of IDLE within the password (which is 4) times the number of possible permutations of the remaining 4 characters (which is 4!).



    So the probability will be $frac{4(4!)}{8!}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 7 hours ago









    Zaeem HussainZaeem Hussain

    785




    785








    • 1




      $begingroup$
      the number of possible positions of IDLE within the password should be 5.
      $endgroup$
      – Julius
      5 hours ago














    • 1




      $begingroup$
      the number of possible positions of IDLE within the password should be 5.
      $endgroup$
      – Julius
      5 hours ago








    1




    1




    $begingroup$
    the number of possible positions of IDLE within the password should be 5.
    $endgroup$
    – Julius
    5 hours ago




    $begingroup$
    the number of possible positions of IDLE within the password should be 5.
    $endgroup$
    – Julius
    5 hours ago











    1












    $begingroup$

    Your question can be interpreted in two different ways. The way of choosing the password as a random permutation of "IDLE1987" was dealt in the answer by Zaeem Hussein. Here I deal with the other interpretation, where the letters of the password are chosen independently and uniformly from $8$ available ones.



    Suppose, we generated a password and a subword "IDLE" appeared in it. If the password has length 8, then the letter 'I' of "IDLE" is either first in the passport, or second, or third, or fourth ($4$ different cases). In any of those cases only it and the $3$ letters after it matter, so the probability of each of these situations is $frac{1}{8^4}$. However, one does not simply sum them up in order to get the answer, as then we will count one possible password twice. That is "IDLEIDLE". Its probability is $frac{1}{8^8}$. So the final probability is $frac{4}{8^4} - frac{1}{8^8} = frac{16383}{16777216}$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Your question can be interpreted in two different ways. The way of choosing the password as a random permutation of "IDLE1987" was dealt in the answer by Zaeem Hussein. Here I deal with the other interpretation, where the letters of the password are chosen independently and uniformly from $8$ available ones.



      Suppose, we generated a password and a subword "IDLE" appeared in it. If the password has length 8, then the letter 'I' of "IDLE" is either first in the passport, or second, or third, or fourth ($4$ different cases). In any of those cases only it and the $3$ letters after it matter, so the probability of each of these situations is $frac{1}{8^4}$. However, one does not simply sum them up in order to get the answer, as then we will count one possible password twice. That is "IDLEIDLE". Its probability is $frac{1}{8^8}$. So the final probability is $frac{4}{8^4} - frac{1}{8^8} = frac{16383}{16777216}$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Your question can be interpreted in two different ways. The way of choosing the password as a random permutation of "IDLE1987" was dealt in the answer by Zaeem Hussein. Here I deal with the other interpretation, where the letters of the password are chosen independently and uniformly from $8$ available ones.



        Suppose, we generated a password and a subword "IDLE" appeared in it. If the password has length 8, then the letter 'I' of "IDLE" is either first in the passport, or second, or third, or fourth ($4$ different cases). In any of those cases only it and the $3$ letters after it matter, so the probability of each of these situations is $frac{1}{8^4}$. However, one does not simply sum them up in order to get the answer, as then we will count one possible password twice. That is "IDLEIDLE". Its probability is $frac{1}{8^8}$. So the final probability is $frac{4}{8^4} - frac{1}{8^8} = frac{16383}{16777216}$






        share|cite|improve this answer











        $endgroup$



        Your question can be interpreted in two different ways. The way of choosing the password as a random permutation of "IDLE1987" was dealt in the answer by Zaeem Hussein. Here I deal with the other interpretation, where the letters of the password are chosen independently and uniformly from $8$ available ones.



        Suppose, we generated a password and a subword "IDLE" appeared in it. If the password has length 8, then the letter 'I' of "IDLE" is either first in the passport, or second, or third, or fourth ($4$ different cases). In any of those cases only it and the $3$ letters after it matter, so the probability of each of these situations is $frac{1}{8^4}$. However, one does not simply sum them up in order to get the answer, as then we will count one possible password twice. That is "IDLEIDLE". Its probability is $frac{1}{8^8}$. So the final probability is $frac{4}{8^4} - frac{1}{8^8} = frac{16383}{16777216}$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 7 hours ago

























        answered 7 hours ago









        Yanior WegYanior Weg

        1,76311138




        1,76311138






















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