If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What is the probability of an event happening in some interval given probability of it in x interval?what is the probability that they getting on the bus?probability question shooting starThree machines are in use certain number of minutes each hour. What is the probability that at least one will be in use at a given moment of the day?What will the probability of scoring prime no. of goals in a game?Probability of takeover announcement in the next hourProbability of an event occurring within a smaller time interval if one knows the probability of occurrence over a larger time intervalSimple probability calculationsWhat is the probability that some random event won't happen in the next 10 minutes given it happened exactly twice in the last 120 minutes?Find the total probability of at least one take-off distaster happening over one yearSolving simple Probability with variational inference

Does the Mueller report show a conspiracy between Russia and the Trump Campaign?

What is the difference between a "ranged attack" and a "ranged weapon attack"?

What is the chair depicted in Cesare Maccari's 1889 painting "Cicerone denuncia Catilina"?

What does 丫 mean? 丫是什么意思?

How to write capital alpha?

How could we fake a moon landing now?

How often does castling occur in grandmaster games?

What initially awakened the Balrog?

Would it be easier to apply for a UK visa if there is a host family to sponsor for you in going there?

Time evolution of a Gaussian wave packet, why convert to k-space?

How did Fremen produce and carry enough thumpers to use Sandworms as de facto Ubers?

What does this say in Elvish?

Google .dev domain strangely redirects to https

How to report t statistic from R

Is there any word for a place full of confusion?

Karn the great creator - 'card from outside the game' in sealed

How would a mousetrap for use in space work?

How do living politicians protect their readily obtainable signatures from misuse?

What is the meaning of 'breadth' in breadth first search?

Can the Flaming Sphere spell be rammed into multiple Tiny creatures that are in the same 5-foot square?

What's the difference between the capability remove_users and delete_users?

Co-worker has annoying ringtone

Significance of Cersei's obsession with elephants?

Most bit efficient text communication method?



If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What is the probability of an event happening in some interval given probability of it in x interval?what is the probability that they getting on the bus?probability question shooting starThree machines are in use certain number of minutes each hour. What is the probability that at least one will be in use at a given moment of the day?What will the probability of scoring prime no. of goals in a game?Probability of takeover announcement in the next hourProbability of an event occurring within a smaller time interval if one knows the probability of occurrence over a larger time intervalSimple probability calculationsWhat is the probability that some random event won't happen in the next 10 minutes given it happened exactly twice in the last 120 minutes?Find the total probability of at least one take-off distaster happening over one yearSolving simple Probability with variational inference










3












$begingroup$


Poorly worded title but I don't know what the nature of this probability question is called.



I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?



Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:



The chance of the dog not barking in a given hour is 1-84% = 16%



If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.



Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.



Question1: Is this correct?



Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:



p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.



Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.



91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.



What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    This related question may be useful: math.stackexchange.com/questions/1376785/…
    $endgroup$
    – 雨が好きな人
    6 hours ago















3












$begingroup$


Poorly worded title but I don't know what the nature of this probability question is called.



I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?



Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:



The chance of the dog not barking in a given hour is 1-84% = 16%



If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.



Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.



Question1: Is this correct?



Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:



p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.



Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.



91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.



What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    This related question may be useful: math.stackexchange.com/questions/1376785/…
    $endgroup$
    – 雨が好きな人
    6 hours ago













3












3








3





$begingroup$


Poorly worded title but I don't know what the nature of this probability question is called.



I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?



Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:



The chance of the dog not barking in a given hour is 1-84% = 16%



If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.



Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.



Question1: Is this correct?



Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:



p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.



Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.



91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.



What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?










share|cite|improve this question











$endgroup$




Poorly worded title but I don't know what the nature of this probability question is called.



I was asked a question:
If the probability of a dog barking one or more times in a given hour is 84%, then what is the probability of a dog barking in 30 minutes?



Since I was told that the first solution I wanted to jump to of 42% is incorrect, I was then presented with the following steps:



The chance of the dog not barking in a given hour is 1-84% = 16%



If the chance of a dog not barking over the course of 2 units - 2 half hours for a total of one hour, then x * x = 16%.



Thus, the probability that the does does NOT bark in 30 minutes is $sqrt16%$ = 40%. Therefore, the probability of the dog barking in a given 30 minutes is 1-40% = 60%.



Question1: Is this correct?



Question2: Rather than work with the inverse probability 16%, surely I can apply the same logic with just 84% and arrive at the same answer? So, if the probability of the dog barking one or more times in an hour is 84%, then this 84% could also be represented as the probability of two 30 minute instances of a dog barking at least once in each instance. In that case:



p(dog barks in 1st half hour AND dog barks in second half hour) = 84%.



Thus the chance of the dog barking in the first half hour is is $sqrt0.84$ = 91.65%.



91.65% does not equal 60% which is what I arrived at by going the negative probability route. I was expecting both numbers to match.



What is the correct way to calculate the probability of a dog barking over 30 minutes if we know that the probability of the dog barking over an hour is 84%?







probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 54 mins ago









YuiTo Cheng

2,58841037




2,58841037










asked 7 hours ago









Doug FirDoug Fir

4688




4688







  • 1




    $begingroup$
    This related question may be useful: math.stackexchange.com/questions/1376785/…
    $endgroup$
    – 雨が好きな人
    6 hours ago












  • 1




    $begingroup$
    This related question may be useful: math.stackexchange.com/questions/1376785/…
    $endgroup$
    – 雨が好きな人
    6 hours ago







1




1




$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
6 hours ago




$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
6 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.



However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.



    We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.



    If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.






    share|cite|improve this answer








    New contributor




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






    $endgroup$




















      2












      $begingroup$

      The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.



      Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.






      share|cite|improve this answer









      $endgroup$













        Your Answer








        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
        );



        );













        draft saved

        draft discarded


















        StackExchange.ready(
        function ()
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3194064%2fif-the-probability-of-a-dog-barking-one-or-more-times-in-a-given-hour-is-84-th%23new-answer', 'question_page');

        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.



        However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.






        share|cite|improve this answer









        $endgroup$

















          2












          $begingroup$

          The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.



          However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.






          share|cite|improve this answer









          $endgroup$















            2












            2








            2





            $begingroup$

            The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.



            However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.






            share|cite|improve this answer









            $endgroup$



            The answer of 60% relies on the assumption that the event of the dog barking in the first half hour is independent of the dog barking in the second half hour. This is not necessarily true unless it is given (e.g., if the mailman comes exactly once in that hour, and the dog always barks when the mailman comes...). It also assumes the probability of the dog barking is the same in each half hour, which is again not necessarily true unless it is given.



            However, given the assumptions above, the 60% answer is correct. You cannot apply the reasoning directly to the 84% probability though. The key is this: the dog does not bark in the whole hour if and only if the dog does not bark in both half-hour intervals. But it is not true that the dog barks during the whole hour if and only if it barks in both half-hour intervals - it need only bark in one of them.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 6 hours ago









            kccukccu

            11.5k11231




            11.5k11231





















                2












                $begingroup$

                The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.



                We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.



                If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.






                share|cite|improve this answer








                New contributor




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






                $endgroup$

















                  2












                  $begingroup$

                  The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.



                  We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.



                  If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.






                  share|cite|improve this answer








                  New contributor




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






                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.



                    We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.



                    If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.






                    share|cite|improve this answer








                    New contributor




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






                    $endgroup$



                    The basic idea here is that the probability of a dog barking in any given window of time should exist and be independent of everything else. Thus, if you want to talk about the probability of a dog not barking in a 1 hour window, this will be exactly equivalent to a dog not barking in two consecutive 30 minute windows, or a dog not barking in 4 consecutive 15 minute windows, and so on.



                    We were told that the probability of a dog barking at least once during an hour is $0.84$. As you have noted, by working with the complementary event -- the dog not barking -- you can arrive at the correct answer. If the dog has a probability of not barking in 1 hour of $0.16$, and this is equivalent to a dog not barking barking in two consecutive 30 minute windows (which has a probability of say, $x$), then since everything in sight is completely independent we know that $x^2=0.16$, or that $x=.4$. But that was the probability that a dog doesn't bark, so the complementary probability is $0.6$.



                    If you don't want to use complementary events (which you should really really want to use since they make life easy in a lot of ways), how could you go about getting this answer "directly"? Well, call the probability of the dog barking in a 30 minute window $y$. How many ways can we have a dog bark at least once in a full hour? Well, it could not bark in the first half, then bark in the second; or it could bark in the first, and not in the second; or it could bark in both halves. This would be $(1-y)y+y(1-y)+y^2=.84$. If you solve that you will find that $y=0.6$, as we expected.







                    share|cite|improve this answer








                    New contributor




                    ItsJustLogsBro 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 answer



                    share|cite|improve this answer






                    New contributor




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









                    answered 6 hours ago









                    ItsJustLogsBroItsJustLogsBro

                    461




                    461




                    New contributor




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





                    New contributor





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






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





















                        2












                        $begingroup$

                        The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.



                        Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.






                        share|cite|improve this answer









                        $endgroup$

















                          2












                          $begingroup$

                          The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.



                          Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.






                          share|cite|improve this answer









                          $endgroup$















                            2












                            2








                            2





                            $begingroup$

                            The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.



                            Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.






                            share|cite|improve this answer









                            $endgroup$



                            The analysis leading to 60% is correct. Your analysis does not take into account possibility of two barks in the same half hour, but none in the other.



                            Also both attempts at analysis assumes barks are independent. I doubt if most dogs behave that way.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 6 hours ago









                            herb steinbergherb steinberg

                            3,2282311




                            3,2282311



























                                draft saved

                                draft discarded
















































                                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.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3194064%2fif-the-probability-of-a-dog-barking-one-or-more-times-in-a-given-hour-is-84-th%23new-answer', 'question_page');

                                );

                                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







                                Popular posts from this blog

                                名間水力發電廠 目录 沿革 設施 鄰近設施 註釋 外部連結 导航菜单23°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.7113923°50′10″N 120°42′41″E / 23.83611°N 120.71139°E / 23.83611; 120.71139計畫概要原始内容臺灣第一座BOT 模式開發的水力發電廠-名間水力電廠名間水力發電廠 水利署首件BOT案原始内容《小檔案》名間電廠 首座BOT水力發電廠原始内容名間電廠BOT - 經濟部水利署中區水資源局

                                Prove that NP is closed under karp reduction?Space(n) not closed under Karp reductions - what about NTime(n)?Class P is closed under rotation?Prove or disprove that $NL$ is closed under polynomial many-one reductions$mathbfNC_2$ is closed under log-space reductionOn Karp reductionwhen can I know if a class (complexity) is closed under reduction (cook/karp)Check if class $PSPACE$ is closed under polyonomially space reductionIs NPSPACE also closed under polynomial-time reduction and under log-space reduction?Prove PSPACE is closed under complement?Prove PSPACE is closed under union?

                                Is my guitar’s action too high? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Strings too stiff on a recently purchased acoustic guitar | Cort AD880CEIs the action of my guitar really high?Μy little finger is too weak to play guitarWith guitar, how long should I give my fingers to strengthen / callous?When playing a fret the guitar sounds mutedPlaying (Barre) chords up the guitar neckI think my guitar strings are wound too tight and I can't play barre chordsF barre chord on an SG guitarHow to find to the right strings of a barre chord by feel?High action on higher fret on my steel acoustic guitar