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 23:30 UTC (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

What would you call this weird metallic apparatus that allows you to lift people?

Random body shuffle every night—can we still function?

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

What's the point of the test set?

Is there public access to the Meteor Crater in Arizona?

How does Belgium enforce obligatory attendance in elections?

How to compare two different files line by line in unix?

Why does 14 CFR have skipped subparts in my ASA 2019 FAR/AIM book?

How often does castling occur in grandmaster games?

A letter with no particular backstory

Tannaka duality for semisimple groups

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

What to do with repeated rejections for phd position

Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?

Did any compiler fully use 80-bit floating point?

Can a Beast Master ranger change beast companions?

Deconstruction is ambiguous

How to identify unknown coordinate type and convert to lat/lon?

What does Turing mean by this statement?

How do I find out the mythology and history of my Fortress?

Most bit efficient text communication method?

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

Google .dev domain strangely redirects to https

Do wooden building fires get hotter than 600°C?



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 23:30 UTC (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%?










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
    3 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
    3 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










asked 3 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
    3 hours ago












  • 1




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







1




1




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




$begingroup$
This related question may be useful: math.stackexchange.com/questions/1376785/…
$endgroup$
– 雨が好きな人
3 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%2f1-probability-to-calculate-two-events-in-a-row%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 3 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 3 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 3 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%2f1-probability-to-calculate-two-events-in-a-row%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

                                Oświęcim Innehåll Historia | Källor | Externa länkar | Navigeringsmeny50°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.2213950°2′18″N 19°13′17″Ö / 50.03833°N 19.22139°Ö / 50.03833; 19.221393089658Nordisk familjebok, AuschwitzInsidan tro och existensJewish Community i OświęcimAuschwitz Jewish Center: MuseumAuschwitz Jewish Center

                                Valle di Casies Indice Geografia fisica | Origini del nome | Storia | Società | Amministrazione | Sport | Note | Bibliografia | Voci correlate | Altri progetti | Collegamenti esterni | Menu di navigazione46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)46°46′N 12°11′E / 46.766667°N 12.183333°E46.766667; 12.183333 (Valle di Casies)Sito istituzionaleAstat Censimento della popolazione 2011 - Determinazione della consistenza dei tre gruppi linguistici della Provincia Autonoma di Bolzano-Alto Adige - giugno 2012Numeri e fattiValle di CasiesDato IstatTabella dei gradi/giorno dei Comuni italiani raggruppati per Regione e Provincia26 agosto 1993, n. 412Heraldry of the World: GsiesStatistiche I.StatValCasies.comWikimedia CommonsWikimedia CommonsValle di CasiesSito ufficialeValle di CasiesMM14870458910042978-6

                                Typsetting diagram chases (with TikZ?) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to define the default vertical distance between nodes?Draw edge on arcNumerical conditional within tikz keys?TikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of themHow to place nodes in an absolute coordinate system in tikzCommutative diagram with curve connecting between nodesTikz with standalone: pinning tikz coordinates to page cmDrawing a Decision Diagram with Tikz and layout manager