Probability of 8 bit binary string whose first three bits are 001, 010, or 100.Probability - Combinations and permutationsProbability of choosing two equal bits from three random bitsA probabilty of error calculationBitstring ProbabilityProbability problem involving random stringsFlawed understanding of independent eventsCombined probability of hit in look up tables with some common index bitsHelp with total probability of a transmitted message (Total probability?)expected error when flipping bits of binary stringConfused with probability
How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?
What are the G forces leaving Earth orbit?
Processor speed limited at 0.4 Ghz
Unlock My Phone! February 2018
Is it possible to map the firing of neurons in the human brain so as to stimulate artificial memories in someone else?
Was the Stack Exchange "Happy April Fools" page fitting with the '90's code?
How do I exit BASH while loop using modulus operator?
Getting extremely large arrows with tikzcd
Mathematica command that allows it to read my intentions
Car headlights in a world without electricity
In the UK, is it possible to get a referendum by a court decision?
How dangerous is XSS
How badly should I try to prevent a user from XSSing themselves?
Why were 5.25" floppy drives cheaper than 8"?
What's the meaning of "Sollensaussagen"?
Why was Sir Cadogan fired?
Where would I need my direct neural interface to be implanted?
What is the fastest integer factorization to break RSA?
How to prevent "they're falling in love" trope
Notepad++ delete until colon for every line with replace all
Is this draw by repetition?
Can a virus destroy the BIOS of a modern computer?
Can someone clarify Hamming's notion of important problems in relation to modern academia?
Placement of More Information/Help Icon button for Radio Buttons
Probability of 8 bit binary string whose first three bits are 001, 010, or 100.
Probability - Combinations and permutationsProbability of choosing two equal bits from three random bitsA probabilty of error calculationBitstring ProbabilityProbability problem involving random stringsFlawed understanding of independent eventsCombined probability of hit in look up tables with some common index bitsHelp with total probability of a transmitted message (Total probability?)expected error when flipping bits of binary stringConfused with probability
$begingroup$
A bit unsure of how to approach the following question.
Consider the uniform probability space where the set of outcomes consists of all 8-bit binary strings. What is the probability of each of the following events?
a) The first three bits of the string are 001, 010, or 100.
I calculated the total number of 8 bit string sequences to be 2^8 or 256.
Now, do I calculate the probability of the first three digits being a particular sequence in the following way :
Pr["001"] = .5 x .5 x .5 = .125 or 32 sequences
Pr["010"] = .5 x .5 x .5 = .125 or 32 sequesces
Pr["111"] = .5 x .5 x .5 = .125 or 32 sequences
Pr[001 or 010 or 111] = (32 + 32 + 32)/256 = .375
I feel I might be doing something wrong. Any help would be greatly appreciated!
probability
asked Mar 20 at 18:18
RobinRobin
525
$endgroup$
add a comment |
$begingroup$
A bit unsure of how to approach the following question.
Consider the uniform probability space where the set of outcomes consists of all 8-bit binary strings. What is the probability of each of the following events?
a) The first three bits of the string are 001, 010, or 100.
I calculated the total number of 8 bit string sequences to be 2^8 or 256.
Now, do I calculate the probability of the first three digits being a particular sequence in the following way :
Pr["001"] = .5 x .5 x .5 = .125 or 32 sequences
Pr["010"] = .5 x .5 x .5 = .125 or 32 sequesces
Pr["111"] = .5 x .5 x .5 = .125 or 32 sequences
Pr[001 or 010 or 111] = (32 + 32 + 32)/256 = .375
I feel I might be doing something wrong. Any help would be greatly appreciated!
probability
asked Mar 20 at 18:18
RobinRobin
525
$endgroup$
5
$begingroup$
You are correct, but you could have done this problem a lot more easily. You note that each bit is independent of the next, so you can just take the space of all 3 bit strings, and find the probability that a given one is one of the three you've chosen. That immediately gives you $frac38$.
$endgroup$
– Don Thousand
Mar 20 at 18:20
add a comment |
$begingroup$
A bit unsure of how to approach the following question.
Consider the uniform probability space where the set of outcomes consists of all 8-bit binary strings. What is the probability of each of the following events?
a) The first three bits of the string are 001, 010, or 100.
I calculated the total number of 8 bit string sequences to be 2^8 or 256.
Now, do I calculate the probability of the first three digits being a particular sequence in the following way :
Pr["001"] = .5 x .5 x .5 = .125 or 32 sequences
Pr["010"] = .5 x .5 x .5 = .125 or 32 sequesces
Pr["111"] = .5 x .5 x .5 = .125 or 32 sequences
Pr[001 or 010 or 111] = (32 + 32 + 32)/256 = .375
I feel I might be doing something wrong. Any help would be greatly appreciated!
probability
asked Mar 20 at 18:18
RobinRobin
525
$endgroup$
A bit unsure of how to approach the following question.
Consider the uniform probability space where the set of outcomes consists of all 8-bit binary strings. What is the probability of each of the following events?
a) The first three bits of the string are 001, 010, or 100.
I calculated the total number of 8 bit string sequences to be 2^8 or 256.
Now, do I calculate the probability of the first three digits being a particular sequence in the following way :
Pr["001"] = .5 x .5 x .5 = .125 or 32 sequences
Pr["010"] = .5 x .5 x .5 = .125 or 32 sequesces
Pr["111"] = .5 x .5 x .5 = .125 or 32 sequences
Pr[001 or 010 or 111] = (32 + 32 + 32)/256 = .375
I feel I might be doing something wrong. Any help would be greatly appreciated!
probability
probability
asked Mar 20 at 18:18
RobinRobin
525
asked Mar 20 at 18:18
RobinRobin
525
asked Mar 20 at 18:18
RobinRobin
525
asked Mar 20 at 18:18
RobinRobin
525
asked Mar 20 at 18:18
RobinRobin
525
525
5
$begingroup$
You are correct, but you could have done this problem a lot more easily. You note that each bit is independent of the next, so you can just take the space of all 3 bit strings, and find the probability that a given one is one of the three you've chosen. That immediately gives you $frac38$.
$endgroup$
– Don Thousand
Mar 20 at 18:20
add a comment |
5
$begingroup$
You are correct, but you could have done this problem a lot more easily. You note that each bit is independent of the next, so you can just take the space of all 3 bit strings, and find the probability that a given one is one of the three you've chosen. That immediately gives you $frac38$.
$endgroup$
– Don Thousand
Mar 20 at 18:20
5
5
$begingroup$
You are correct, but you could have done this problem a lot more easily. You note that each bit is independent of the next, so you can just take the space of all 3 bit strings, and find the probability that a given one is one of the three you've chosen. That immediately gives you $frac38$.
$endgroup$
– Don Thousand
Mar 20 at 18:20
$begingroup$
You are correct, but you could have done this problem a lot more easily. You note that each bit is independent of the next, so you can just take the space of all 3 bit strings, and find the probability that a given one is one of the three you've chosen. That immediately gives you $frac38$.
$endgroup$
– Don Thousand
Mar 20 at 18:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have three chances of getting the first three bits with a single one out of the $2^3$ possibilities. That's $frac38$. Then we have $2^5$ combinations for the remaining five bits but they don't matter because, for every value they may have, there is still the $frac38$ chance that the leading digits will have one one. Answer: $frac38= 0.375$ so you got it right.
answered Mar 20 at 18:42
poetasispoetasis
432317
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155823%2fprobability-of-8-bit-binary-string-whose-first-three-bits-are-001-010-or-100%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have three chances of getting the first three bits with a single one out of the $2^3$ possibilities. That's $frac38$. Then we have $2^5$ combinations for the remaining five bits but they don't matter because, for every value they may have, there is still the $frac38$ chance that the leading digits will have one one. Answer: $frac38= 0.375$ so you got it right.
answered Mar 20 at 18:42
poetasispoetasis
432317
$endgroup$
add a comment |
$begingroup$
We have three chances of getting the first three bits with a single one out of the $2^3$ possibilities. That's $frac38$. Then we have $2^5$ combinations for the remaining five bits but they don't matter because, for every value they may have, there is still the $frac38$ chance that the leading digits will have one one. Answer: $frac38= 0.375$ so you got it right.
answered Mar 20 at 18:42
poetasispoetasis
432317
$endgroup$
add a comment |
$begingroup$
We have three chances of getting the first three bits with a single one out of the $2^3$ possibilities. That's $frac38$. Then we have $2^5$ combinations for the remaining five bits but they don't matter because, for every value they may have, there is still the $frac38$ chance that the leading digits will have one one. Answer: $frac38= 0.375$ so you got it right.
answered Mar 20 at 18:42
poetasispoetasis
432317
$endgroup$
We have three chances of getting the first three bits with a single one out of the $2^3$ possibilities. That's $frac38$. Then we have $2^5$ combinations for the remaining five bits but they don't matter because, for every value they may have, there is still the $frac38$ chance that the leading digits will have one one. Answer: $frac38= 0.375$ so you got it right.
answered Mar 20 at 18:42
poetasispoetasis
432317
edited Mar 21 at 18:02
answered Mar 20 at 18:42
poetasispoetasis
432317
answered Mar 20 at 18:42
poetasispoetasis
432317
answered Mar 20 at 18:42
poetasispoetasis
432317
432317
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155823%2fprobability-of-8-bit-binary-string-whose-first-three-bits-are-001-010-or-100%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
5
$begingroup$
You are correct, but you could have done this problem a lot more easily. You note that each bit is independent of the next, so you can just take the space of all 3 bit strings, and find the probability that a given one is one of the three you've chosen. That immediately gives you $frac38$.
$endgroup$
– Don Thousand
Mar 20 at 18:20