Interpretation of a seemingly simple probability question

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I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:



There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?



What is your answer, and why? Thanks in advance.







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    up vote
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    I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:



    There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?



    What is your answer, and why? Thanks in advance.







    share|cite|improve this question







    New contributor




    JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
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      1





      I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:



      There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?



      What is your answer, and why? Thanks in advance.







      share|cite|improve this question







      New contributor




      JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:



      There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?



      What is your answer, and why? Thanks in advance.









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      JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked Sep 6 at 18:22









      JPMaverick

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          3 Answers
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          It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.



          enter image description here



          Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.






          share|cite|improve this answer






















          • Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
            – JPMaverick
            Sep 6 at 18:59










          • The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
            – Nuclear Wang
            Sep 6 at 19:29










          • Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
            – JPMaverick
            Sep 6 at 19:36











          • I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
            – JPMaverick
            Sep 6 at 19:52






          • 1




            @JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
            – Nuclear Wang
            Sep 6 at 19:59


















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          Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
          $$
          fracacdot b2frac12a cdot 2b = frac18.
          $$






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            Stumbled across this and it got in my head. :-)



            The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.



            For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.



            Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.



            • The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)

            • The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$

            • etc.

            Putting each of those terms into a summation, we get:



            $(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$



            Or,



            $sum_i=0^n-1 fracin^2$



            How to handle the fact A and B do not deliver the same number of packages?



            I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.



            How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.



            $frac12 cdot sum_i=0^b-1 fracib^2$






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            • The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
              – whuber♦
              yesterday











            • You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
              – Tommy
              yesterday










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            3 Answers
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            3 Answers
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            It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.



            enter image description here



            Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.






            share|cite|improve this answer






















            • Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
              – JPMaverick
              Sep 6 at 18:59










            • The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
              – Nuclear Wang
              Sep 6 at 19:29










            • Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
              – JPMaverick
              Sep 6 at 19:36











            • I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
              – JPMaverick
              Sep 6 at 19:52






            • 1




              @JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
              – Nuclear Wang
              Sep 6 at 19:59















            up vote
            8
            down vote













            It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.



            enter image description here



            Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.






            share|cite|improve this answer






















            • Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
              – JPMaverick
              Sep 6 at 18:59










            • The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
              – Nuclear Wang
              Sep 6 at 19:29










            • Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
              – JPMaverick
              Sep 6 at 19:36











            • I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
              – JPMaverick
              Sep 6 at 19:52






            • 1




              @JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
              – Nuclear Wang
              Sep 6 at 19:59













            up vote
            8
            down vote










            up vote
            8
            down vote









            It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.



            enter image description here



            Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.






            share|cite|improve this answer














            It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.



            enter image description here



            Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 6 at 18:52

























            answered Sep 6 at 18:47









            Nuclear Wang

            2,239818




            2,239818











            • Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
              – JPMaverick
              Sep 6 at 18:59










            • The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
              – Nuclear Wang
              Sep 6 at 19:29










            • Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
              – JPMaverick
              Sep 6 at 19:36











            • I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
              – JPMaverick
              Sep 6 at 19:52






            • 1




              @JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
              – Nuclear Wang
              Sep 6 at 19:59

















            • Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
              – JPMaverick
              Sep 6 at 18:59










            • The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
              – Nuclear Wang
              Sep 6 at 19:29










            • Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
              – JPMaverick
              Sep 6 at 19:36











            • I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
              – JPMaverick
              Sep 6 at 19:52






            • 1




              @JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
              – Nuclear Wang
              Sep 6 at 19:59
















            Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
            – JPMaverick
            Sep 6 at 18:59




            Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
            – JPMaverick
            Sep 6 at 18:59












            The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
            – Nuclear Wang
            Sep 6 at 19:29




            The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
            – Nuclear Wang
            Sep 6 at 19:29












            Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
            – JPMaverick
            Sep 6 at 19:36





            Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
            – JPMaverick
            Sep 6 at 19:36













            I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
            – JPMaverick
            Sep 6 at 19:52




            I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
            – JPMaverick
            Sep 6 at 19:52




            1




            1




            @JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
            – Nuclear Wang
            Sep 6 at 19:59





            @JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
            – Nuclear Wang
            Sep 6 at 19:59













            up vote
            3
            down vote













            Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
            $$
            fracacdot b2frac12a cdot 2b = frac18.
            $$






            share|cite|improve this answer








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            katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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              3
              down vote













              Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
              $$
              fracacdot b2frac12a cdot 2b = frac18.
              $$






              share|cite|improve this answer








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                up vote
                3
                down vote










                up vote
                3
                down vote









                Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
                $$
                fracacdot b2frac12a cdot 2b = frac18.
                $$






                share|cite|improve this answer








                New contributor




                katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
                $$
                fracacdot b2frac12a cdot 2b = frac18.
                $$







                share|cite|improve this answer








                New contributor




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                share|cite|improve this answer



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                answered Sep 6 at 22:33









                katosh

                1093




                1093




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                    up vote
                    0
                    down vote













                    Stumbled across this and it got in my head. :-)



                    The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.



                    For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.



                    Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.



                    • The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)

                    • The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$

                    • etc.

                    Putting each of those terms into a summation, we get:



                    $(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$



                    Or,



                    $sum_i=0^n-1 fracin^2$



                    How to handle the fact A and B do not deliver the same number of packages?



                    I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.



                    How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.



                    $frac12 cdot sum_i=0^b-1 fracib^2$






                    share|cite|improve this answer










                    New contributor




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

















                    • The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
                      – whuber♦
                      yesterday











                    • You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
                      – Tommy
                      yesterday














                    up vote
                    0
                    down vote













                    Stumbled across this and it got in my head. :-)



                    The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.



                    For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.



                    Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.



                    • The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)

                    • The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$

                    • etc.

                    Putting each of those terms into a summation, we get:



                    $(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$



                    Or,



                    $sum_i=0^n-1 fracin^2$



                    How to handle the fact A and B do not deliver the same number of packages?



                    I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.



                    How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.



                    $frac12 cdot sum_i=0^b-1 fracib^2$






                    share|cite|improve this answer










                    New contributor




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

















                    • The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
                      – whuber♦
                      yesterday











                    • You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
                      – Tommy
                      yesterday












                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Stumbled across this and it got in my head. :-)



                    The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.



                    For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.



                    Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.



                    • The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)

                    • The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$

                    • etc.

                    Putting each of those terms into a summation, we get:



                    $(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$



                    Or,



                    $sum_i=0^n-1 fracin^2$



                    How to handle the fact A and B do not deliver the same number of packages?



                    I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.



                    How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.



                    $frac12 cdot sum_i=0^b-1 fracib^2$






                    share|cite|improve this answer










                    New contributor




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









                    Stumbled across this and it got in my head. :-)



                    The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.



                    For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.



                    Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.



                    • The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)

                    • The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$

                    • etc.

                    Putting each of those terms into a summation, we get:



                    $(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$



                    Or,



                    $sum_i=0^n-1 fracin^2$



                    How to handle the fact A and B do not deliver the same number of packages?



                    I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.



                    How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.



                    $frac12 cdot sum_i=0^b-1 fracib^2$







                    share|cite|improve this answer










                    New contributor




                    Tommy 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








                    edited 13 hours ago





















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                    answered yesterday









                    Tommy

                    11




                    11




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                    Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    • The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
                      – whuber♦
                      yesterday











                    • You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
                      – Tommy
                      yesterday
















                    • The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
                      – whuber♦
                      yesterday











                    • You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
                      – Tommy
                      yesterday















                    The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
                    – whuber♦
                    yesterday





                    The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
                    – whuber♦
                    yesterday













                    You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
                    – Tommy
                    yesterday




                    You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
                    – Tommy
                    yesterday










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