4 letters and 4 numbers to choose from, whats the probability that a specific 4 letter word is chosen...
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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
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
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closed as off-topic by GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517 Feb 18 at 23:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
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
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
$endgroup$
closed as off-topic by GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517 Feb 18 at 23:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517
If this question can be reworded to fit the rules in the help center, please edit the question.
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Can a character be used more than once?
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– saulspatz
Feb 9 at 15:53
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Does "ID1987LE" count or must the four characters be consecutive ?
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– WW1
Feb 9 at 15:54
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Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
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– Sujit Bhattacharyya
Feb 9 at 15:56
add a comment |
$begingroup$
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
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
$endgroup$
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
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
edited Feb 9 at 15:50
GNUSupporter 8964民主女神 地下教會
14k82651
14k82651
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
asked Feb 9 at 15:48
Priscilla96Priscilla96
162
162
closed as off-topic by GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517 Feb 18 at 23:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517 Feb 18 at 23:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, NCh, Cesareo, darij grinberg, user159517
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Can a character be used more than once?
$endgroup$
– saulspatz
Feb 9 at 15:53
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Does "ID1987LE" count or must the four characters be consecutive ?
$endgroup$
– WW1
Feb 9 at 15:54
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Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
$endgroup$
– Sujit Bhattacharyya
Feb 9 at 15:56
add a comment |
$begingroup$
Can a character be used more than once?
$endgroup$
– saulspatz
Feb 9 at 15:53
$begingroup$
Does "ID1987LE" count or must the four characters be consecutive ?
$endgroup$
– WW1
Feb 9 at 15:54
$begingroup$
Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
$endgroup$
– Sujit Bhattacharyya
Feb 9 at 15:56
$begingroup$
Can a character be used more than once?
$endgroup$
– saulspatz
Feb 9 at 15:53
$begingroup$
Can a character be used more than once?
$endgroup$
– saulspatz
Feb 9 at 15:53
$begingroup$
Does "ID1987LE" count or must the four characters be consecutive ?
$endgroup$
– WW1
Feb 9 at 15:54
$begingroup$
Does "ID1987LE" count or must the four characters be consecutive ?
$endgroup$
– WW1
Feb 9 at 15:54
$begingroup$
Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
$endgroup$
– Sujit Bhattacharyya
Feb 9 at 15:56
$begingroup$
Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
$endgroup$
– Sujit Bhattacharyya
Feb 9 at 15:56
add a comment |
2 Answers
2
active
oldest
votes
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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 5) times the number of possible permutations of the remaining 4 characters (which is 4!).
So the probability will be $frac{5(4!)}{8!}$.
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
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3
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the number of possible positions of IDLE within the password should be 5.
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– Julius
Feb 9 at 17:55
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Fixed. Thanks for pointing it out.
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– Zaeem Hussain
Feb 10 at 18:39
add a comment |
$begingroup$
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, or fifth ($5$ 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 $frac{5}{8^4} - frac{1}{8^8} = frac{20479}{16777216}$
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 5) times the number of possible permutations of the remaining 4 characters (which is 4!).
So the probability will be $frac{5(4!)}{8!}$.
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
$endgroup$
3
$begingroup$
the number of possible positions of IDLE within the password should be 5.
$endgroup$
– Julius
Feb 9 at 17:55
$begingroup$
Fixed. Thanks for pointing it out.
$endgroup$
– Zaeem Hussain
Feb 10 at 18:39
add a comment |
$begingroup$
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 5) times the number of possible permutations of the remaining 4 characters (which is 4!).
So the probability will be $frac{5(4!)}{8!}$.
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
$endgroup$
3
$begingroup$
the number of possible positions of IDLE within the password should be 5.
$endgroup$
– Julius
Feb 9 at 17:55
$begingroup$
Fixed. Thanks for pointing it out.
$endgroup$
– Zaeem Hussain
Feb 10 at 18:39
add a comment |
$begingroup$
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 5) times the number of possible permutations of the remaining 4 characters (which is 4!).
So the probability will be $frac{5(4!)}{8!}$.
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
$endgroup$
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 5) times the number of possible permutations of the remaining 4 characters (which is 4!).
So the probability will be $frac{5(4!)}{8!}$.
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
edited Feb 10 at 18:37
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
answered Feb 9 at 15:59
Zaeem HussainZaeem Hussain
1657
1657
3
$begingroup$
the number of possible positions of IDLE within the password should be 5.
$endgroup$
– Julius
Feb 9 at 17:55
$begingroup$
Fixed. Thanks for pointing it out.
$endgroup$
– Zaeem Hussain
Feb 10 at 18:39
add a comment |
3
$begingroup$
the number of possible positions of IDLE within the password should be 5.
$endgroup$
– Julius
Feb 9 at 17:55
$begingroup$
Fixed. Thanks for pointing it out.
$endgroup$
– Zaeem Hussain
Feb 10 at 18:39
3
3
$begingroup$
the number of possible positions of IDLE within the password should be 5.
$endgroup$
– Julius
Feb 9 at 17:55
$begingroup$
the number of possible positions of IDLE within the password should be 5.
$endgroup$
– Julius
Feb 9 at 17:55
$begingroup$
Fixed. Thanks for pointing it out.
$endgroup$
– Zaeem Hussain
Feb 10 at 18:39
$begingroup$
Fixed. Thanks for pointing it out.
$endgroup$
– Zaeem Hussain
Feb 10 at 18:39
add a comment |
$begingroup$
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, or fifth ($5$ 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 $frac{5}{8^4} - frac{1}{8^8} = frac{20479}{16777216}$
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
$endgroup$
add a comment |
$begingroup$
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, or fifth ($5$ 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 $frac{5}{8^4} - frac{1}{8^8} = frac{20479}{16777216}$
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
$endgroup$
add a comment |
$begingroup$
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, or fifth ($5$ 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 $frac{5}{8^4} - frac{1}{8^8} = frac{20479}{16777216}$
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
$endgroup$
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, or fifth ($5$ 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 $frac{5}{8^4} - frac{1}{8^8} = frac{20479}{16777216}$
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
edited Feb 10 at 11:24
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
answered Feb 9 at 16:09
Yanior WegYanior Weg
2,72711446
2,72711446
add a comment |
add a comment |
$begingroup$
Can a character be used more than once?
$endgroup$
– saulspatz
Feb 9 at 15:53
$begingroup$
Does "ID1987LE" count or must the four characters be consecutive ?
$endgroup$
– WW1
Feb 9 at 15:54
$begingroup$
Is the password is "IDLE1987"? I mean do you want the probability of getting the password "IDLE1987" from all possible combinations?
$endgroup$
– Sujit Bhattacharyya
Feb 9 at 15:56