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what is the probability that the ID for the first registered user will start with the letter c and end with letter g
Would this method yield a truly random sequence?Probability and Combinations - Is my answer correct?Probability of a letter being in a four letter word.Confusing probability question, please help meWhat is the probability that if six letters are selected from the alphabet with replacement that the chosen letters can be used to spell ASSESS?Probability question with black and white marblesProbability of creating a password with 2 digits and 6 letterscomputing probability that at least one letter matches with envelopeWhat is the Probability Generating Function (PGF) for the “Matching Problem”?Probability of substring occurring at the end of a string
$begingroup$
User IDs on an old computer system consist of 4 letter sequences from the first 8 letters of the English alphabet: a, b, c, d, e, f, g, h. Note that the same letter can be used any number of times in the 4 letter sequence. When a new user first registers, an ID is randomly generated(all outcomes equally likely) with the only condition that it is different from the IDs of all previously registered users.
a) How many different IDs are possible for the first registered user? (Just to control myself i found 8x8x8x8= 4,096)
b) What is the probability that the ID for the first registered user will start with the letter c and end with the letter g?
probability statistics random-variables random
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
User IDs on an old computer system consist of 4 letter sequences from the first 8 letters of the English alphabet: a, b, c, d, e, f, g, h. Note that the same letter can be used any number of times in the 4 letter sequence. When a new user first registers, an ID is randomly generated(all outcomes equally likely) with the only condition that it is different from the IDs of all previously registered users.
a) How many different IDs are possible for the first registered user? (Just to control myself i found 8x8x8x8= 4,096)
b) What is the probability that the ID for the first registered user will start with the letter c and end with the letter g?
probability statistics random-variables random
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Welcome to MSE. Please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Otherwise this question is likely to be closed via community votes or just be ignored. Also, please click on the tinyeditand use MathJax to properly typeset math expressions.
$endgroup$
– Lee David Chung Lin
Mar 11 at 21:40
add a comment |
$begingroup$
User IDs on an old computer system consist of 4 letter sequences from the first 8 letters of the English alphabet: a, b, c, d, e, f, g, h. Note that the same letter can be used any number of times in the 4 letter sequence. When a new user first registers, an ID is randomly generated(all outcomes equally likely) with the only condition that it is different from the IDs of all previously registered users.
a) How many different IDs are possible for the first registered user? (Just to control myself i found 8x8x8x8= 4,096)
b) What is the probability that the ID for the first registered user will start with the letter c and end with the letter g?
probability statistics random-variables random
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
User IDs on an old computer system consist of 4 letter sequences from the first 8 letters of the English alphabet: a, b, c, d, e, f, g, h. Note that the same letter can be used any number of times in the 4 letter sequence. When a new user first registers, an ID is randomly generated(all outcomes equally likely) with the only condition that it is different from the IDs of all previously registered users.
a) How many different IDs are possible for the first registered user? (Just to control myself i found 8x8x8x8= 4,096)
b) What is the probability that the ID for the first registered user will start with the letter c and end with the letter g?
probability statistics random-variables random
probability statistics random-variables random
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 21:42
Ziya Arslan
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
asked Mar 11 at 21:27
Ziya ArslanZiya Arslan
11
11
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ziya Arslan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Welcome to MSE. Please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Otherwise this question is likely to be closed via community votes or just be ignored. Also, please click on the tinyeditand use MathJax to properly typeset math expressions.
$endgroup$
– Lee David Chung Lin
Mar 11 at 21:40
add a comment |
1
$begingroup$
Welcome to MSE. Please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Otherwise this question is likely to be closed via community votes or just be ignored. Also, please click on the tinyeditand use MathJax to properly typeset math expressions.
$endgroup$
– Lee David Chung Lin
Mar 11 at 21:40
1
1
$begingroup$
Welcome to MSE. Please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Otherwise this question is likely to be closed via community votes or just be ignored. Also, please click on the tiny
edit and use MathJax to properly typeset math expressions.$endgroup$
– Lee David Chung Lin
Mar 11 at 21:40
$begingroup$
Welcome to MSE. Please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Otherwise this question is likely to be closed via community votes or just be ignored. Also, please click on the tiny
edit and use MathJax to properly typeset math expressions.$endgroup$
– Lee David Chung Lin
Mar 11 at 21:40
add a comment |
1 Answer
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$begingroup$
You correctly found that there are $8^4=4096$ possible equally likely ids that could be generated for the user.
Now... if it starts with a $c$ and ends with a $g$... then we have in essence $1$ option for the first character (it must be a c), $8$ options for the second character (it could be any of a,b,...,h), $8$ options again for the third character, and $1$ option for the final character (it must be a g).
Applying the rule of product, by multiplying the number of options available for each position we get the total number of passwords which begin with a $c$ and end with a $g$.
Dividing the result by the number of possible ids (noting that the ids are equally likely to occur) gives us the desired probability.
(Note: the number of "good occurrences" divided by the number of "possible occurrences" giving the probability is only a valid technique in the situation where each possible occurrence is explicitly known to be equally likely to occur. It may not be used in situations outside of that and can and will frequently lead to incorrect answers if this warning is ignored)
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
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$begingroup$
You correctly found that there are $8^4=4096$ possible equally likely ids that could be generated for the user.
Now... if it starts with a $c$ and ends with a $g$... then we have in essence $1$ option for the first character (it must be a c), $8$ options for the second character (it could be any of a,b,...,h), $8$ options again for the third character, and $1$ option for the final character (it must be a g).
Applying the rule of product, by multiplying the number of options available for each position we get the total number of passwords which begin with a $c$ and end with a $g$.
Dividing the result by the number of possible ids (noting that the ids are equally likely to occur) gives us the desired probability.
(Note: the number of "good occurrences" divided by the number of "possible occurrences" giving the probability is only a valid technique in the situation where each possible occurrence is explicitly known to be equally likely to occur. It may not be used in situations outside of that and can and will frequently lead to incorrect answers if this warning is ignored)
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
$endgroup$
add a comment |
$begingroup$
You correctly found that there are $8^4=4096$ possible equally likely ids that could be generated for the user.
Now... if it starts with a $c$ and ends with a $g$... then we have in essence $1$ option for the first character (it must be a c), $8$ options for the second character (it could be any of a,b,...,h), $8$ options again for the third character, and $1$ option for the final character (it must be a g).
Applying the rule of product, by multiplying the number of options available for each position we get the total number of passwords which begin with a $c$ and end with a $g$.
Dividing the result by the number of possible ids (noting that the ids are equally likely to occur) gives us the desired probability.
(Note: the number of "good occurrences" divided by the number of "possible occurrences" giving the probability is only a valid technique in the situation where each possible occurrence is explicitly known to be equally likely to occur. It may not be used in situations outside of that and can and will frequently lead to incorrect answers if this warning is ignored)
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
$endgroup$
add a comment |
$begingroup$
You correctly found that there are $8^4=4096$ possible equally likely ids that could be generated for the user.
Now... if it starts with a $c$ and ends with a $g$... then we have in essence $1$ option for the first character (it must be a c), $8$ options for the second character (it could be any of a,b,...,h), $8$ options again for the third character, and $1$ option for the final character (it must be a g).
Applying the rule of product, by multiplying the number of options available for each position we get the total number of passwords which begin with a $c$ and end with a $g$.
Dividing the result by the number of possible ids (noting that the ids are equally likely to occur) gives us the desired probability.
(Note: the number of "good occurrences" divided by the number of "possible occurrences" giving the probability is only a valid technique in the situation where each possible occurrence is explicitly known to be equally likely to occur. It may not be used in situations outside of that and can and will frequently lead to incorrect answers if this warning is ignored)
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
$endgroup$
You correctly found that there are $8^4=4096$ possible equally likely ids that could be generated for the user.
Now... if it starts with a $c$ and ends with a $g$... then we have in essence $1$ option for the first character (it must be a c), $8$ options for the second character (it could be any of a,b,...,h), $8$ options again for the third character, and $1$ option for the final character (it must be a g).
Applying the rule of product, by multiplying the number of options available for each position we get the total number of passwords which begin with a $c$ and end with a $g$.
Dividing the result by the number of possible ids (noting that the ids are equally likely to occur) gives us the desired probability.
(Note: the number of "good occurrences" divided by the number of "possible occurrences" giving the probability is only a valid technique in the situation where each possible occurrence is explicitly known to be equally likely to occur. It may not be used in situations outside of that and can and will frequently lead to incorrect answers if this warning is ignored)
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
answered Mar 11 at 23:20
JMoravitzJMoravitz
48.5k43988
48.5k43988
add a comment |
add a comment |
Ziya Arslan is a new contributor. Be nice, and check out our Code of Conduct.
Ziya Arslan is a new contributor. Be nice, and check out our Code of Conduct.
Ziya Arslan is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Welcome to MSE. Please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Otherwise this question is likely to be closed via community votes or just be ignored. Also, please click on the tiny
editand use MathJax to properly typeset math expressions.$endgroup$
– Lee David Chung Lin
Mar 11 at 21:40