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1-probability to calculate two events in a row



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30UTC (7:30pm US/Eastern)What is the probability of an event happening in some interval given probability of it in x interval?Probability Question on Waiting Timeprobability question shooting starExponential and Uniform distribution with conditional probabilityTwo poisson processes - no independenceWhat will the probability of scoring prime no. of goals in a game?Probability of takeover announcement in the next hourExponential ProbabilityProbability of an event occurring within a smaller time interval if one knows the probability of occurrence over a larger time intervalSimple probability calculationsSolving 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%?










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    $begingroup$
    This related question may be useful: math.stackexchange.com/questions/1376785/…
    $endgroup$
    – 雨が好きな人
    2 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%?










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$endgroup$







  • 1




    $begingroup$
    This related question may be useful: math.stackexchange.com/questions/1376785/…
    $endgroup$
    – 雨が好きな人
    2 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






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asked 2 hours ago









Doug FirDoug Fir

4688




4688







  • 1




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












  • 1




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







1




1




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




$begingroup$
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3 Answers
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$begingroup$

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.






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    ItsJustLogsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      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.






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        3 Answers
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        3 Answers
        3






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        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 2 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.
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                    answered 2 hours ago









                    ItsJustLogsBroItsJustLogsBro

                    461




                    461




                    New contributor




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                    New contributor





                    ItsJustLogsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    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 2 hours ago









                            herb steinbergherb steinberg

                            3,2282311




                            3,2282311



























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                                Францішак Багушэвіч Змест Сям'я | Біяграфія | Творчасць | Мова Багушэвіча | Ацэнкі дзейнасці | Цікавыя факты | Спадчына | Выбраная бібліяграфія | Ушанаванне памяці | У філатэліі | Зноскі | Літаратура | Спасылкі | НавігацыяЛяхоўскі У. Рупіўся дзеля Бога і людзей: Жыццёвы шлях Лявона Вітан-Дубейкаўскага // Вольскі і Памідораў з песняй пра немца Адвакат, паэт, народны заступнік Ашмянскі веснікВ Минске появится площадь Богушевича и улица Сырокомли, Белорусская деловая газета, 19 июля 2001 г.Айцец беларускай нацыянальнай ідэі паўстаў у бронзе Сяргей Аляксандравіч Адашкевіч (1918, Мінск). 80-я гады. Бюст «Францішак Багушэвіч».Яўген Мікалаевіч Ціхановіч. «Партрэт Францішка Багушэвіча»Мікола Мікалаевіч Купава. «Партрэт зачынальніка новай беларускай літаратуры Францішка Багушэвіча»Уладзімір Іванавіч Мелехаў. На помніку «Змагарам за родную мову» Барэльеф «Францішак Багушэвіч»Памяць пра Багушэвіча на Віленшчыне Страчаная сталіца. Беларускія шыльды на вуліцах Вільні«Krynica». Ideologia i przywódcy białoruskiego katolicyzmuФранцішак БагушэвічТворы на knihi.comТворы Францішка Багушэвіча на bellib.byСодаль Уладзімір. Францішак Багушэвіч на Лідчыне;Луцкевіч Антон. Жыцьцё і творчасьць Фр. Багушэвіча ў успамінах ягоных сучасьнікаў // Запісы Беларускага Навуковага таварыства. Вільня, 1938. Сшытак 1. С. 16-34.Большая российская1188761710000 0000 5537 633Xn9209310021619551927869394п

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                                ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result