The probability of Bus A arriving before Bus B The 2019 Stack Overflow Developer Survey Results Are InExpected time of last bus leftProbability at least one of two buses arrive on timeBus stop independent events expected valueWhat is the expected time you have to wait until the first bus comes?Probabilty of 2 buses or more arriving at a bus s top at the same timeContinuous Probability - Bus ArrivingFred-to-bus and bus-to-bus average timesBus arrival probability…Average time waiting for busBus arrival times and minimum of exponential random variables

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The probability of Bus A arriving before Bus B



The 2019 Stack Overflow Developer Survey Results Are InExpected time of last bus leftProbability at least one of two buses arrive on timeBus stop independent events expected valueWhat is the expected time you have to wait until the first bus comes?Probabilty of 2 buses or more arriving at a bus s top at the same timeContinuous Probability - Bus ArrivingFred-to-bus and bus-to-bus average timesBus arrival probability…Average time waiting for busBus arrival times and minimum of exponential random variables










6












$begingroup$


Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?



My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    Mar 23 at 4:10










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    Mar 23 at 4:45










  • $begingroup$
    As a nitpick, the words "probability" and "odds" are not interchangeable. They are related, yes, but they do not mean the same thing. The probability of picking an ace from a well shuffled standard deck is $frac113$. The odds however are $1:12$ for, or equivalently $12:1$ against. If you only ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(Amid B')$, not $P(A/B')$
    $endgroup$
    – JMoravitz
    Mar 23 at 13:59










  • $begingroup$
    @JMoravitz I missed the syntax there , I'll try not to repeat it.
    $endgroup$
    – Vimath
    Mar 23 at 14:53















6












$begingroup$


Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?



My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    Mar 23 at 4:10










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    Mar 23 at 4:45










  • $begingroup$
    As a nitpick, the words "probability" and "odds" are not interchangeable. They are related, yes, but they do not mean the same thing. The probability of picking an ace from a well shuffled standard deck is $frac113$. The odds however are $1:12$ for, or equivalently $12:1$ against. If you only ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(Amid B')$, not $P(A/B')$
    $endgroup$
    – JMoravitz
    Mar 23 at 13:59










  • $begingroup$
    @JMoravitz I missed the syntax there , I'll try not to repeat it.
    $endgroup$
    – Vimath
    Mar 23 at 14:53













6












6








6





$begingroup$


Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?



My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?










share|cite|improve this question











$endgroup$




Bus A arrives at a random time between 2pm and 4pm, and Bus B arrives at a random time between 3pm and 5pm. What are the odds that Bus A arrives before Bus B?



My understanding is that since Bus B cannot possibly arrive between 2 and 3, we can only talk about the time between 3 and 4 pm, when there is an equal probability for both buses arriving. But in this case, the probability of Bus A arriving before B is 50%. No? What am I missing here? Or I should look at the entire timeline, 2 pm - 5 pm? But then in this case, it is still 50%. Where is my thinking wrong?







probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 3:57







IrinaS

















asked Mar 23 at 3:41









IrinaSIrinaS

312




312







  • 1




    $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    Mar 23 at 4:10










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    Mar 23 at 4:45










  • $begingroup$
    As a nitpick, the words "probability" and "odds" are not interchangeable. They are related, yes, but they do not mean the same thing. The probability of picking an ace from a well shuffled standard deck is $frac113$. The odds however are $1:12$ for, or equivalently $12:1$ against. If you only ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(Amid B')$, not $P(A/B')$
    $endgroup$
    – JMoravitz
    Mar 23 at 13:59










  • $begingroup$
    @JMoravitz I missed the syntax there , I'll try not to repeat it.
    $endgroup$
    – Vimath
    Mar 23 at 14:53












  • 1




    $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    Mar 23 at 4:10










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    Mar 23 at 4:45










  • $begingroup$
    As a nitpick, the words "probability" and "odds" are not interchangeable. They are related, yes, but they do not mean the same thing. The probability of picking an ace from a well shuffled standard deck is $frac113$. The odds however are $1:12$ for, or equivalently $12:1$ against. If you only ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(Amid B')$, not $P(A/B')$
    $endgroup$
    – JMoravitz
    Mar 23 at 13:59










  • $begingroup$
    @JMoravitz I missed the syntax there , I'll try not to repeat it.
    $endgroup$
    – Vimath
    Mar 23 at 14:53







1




1




$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
Mar 23 at 4:10




$begingroup$
Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
$endgroup$
– Vimath
Mar 23 at 4:10












$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
Mar 23 at 4:45




$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
Mar 23 at 4:45












$begingroup$
As a nitpick, the words "probability" and "odds" are not interchangeable. They are related, yes, but they do not mean the same thing. The probability of picking an ace from a well shuffled standard deck is $frac113$. The odds however are $1:12$ for, or equivalently $12:1$ against. If you only ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(Amid B')$, not $P(A/B')$
$endgroup$
– JMoravitz
Mar 23 at 13:59




$begingroup$
As a nitpick, the words "probability" and "odds" are not interchangeable. They are related, yes, but they do not mean the same thing. The probability of picking an ace from a well shuffled standard deck is $frac113$. The odds however are $1:12$ for, or equivalently $12:1$ against. If you only ever want to talk about probabilities, then only use the word probability and avoid using the word odds. Also, @Vimath conditional probabilities are written with a vertical bar, not a slanted bar. It should be $P(Amid B')$, not $P(A/B')$
$endgroup$
– JMoravitz
Mar 23 at 13:59












$begingroup$
@JMoravitz I missed the syntax there , I'll try not to repeat it.
$endgroup$
– Vimath
Mar 23 at 14:53




$begingroup$
@JMoravitz I missed the syntax there , I'll try not to repeat it.
$endgroup$
– Vimath
Mar 23 at 14:53










7 Answers
7






active

oldest

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14












$begingroup$

Guide:



1) Draw rectangle $2le xle 4,$ $3le yle 5$.



2) The area of the rectangle is $4$, so pdf is $1/4$.



3) Draw line $y=x$.



4) Find area of the rectangle above the line, which is $7/2$.



5) Finally, the required probability is $7/2cdot 1/4=7/8$.



Here is the graph:



$hspace2cm$ enter image description here



Bus $A$ arriving before bus $B$:
$$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$
Bus $A$ arriving after bus $B$:
$$E(3.7,3.3).$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
    $endgroup$
    – IrinaS
    Mar 23 at 5:15










  • $begingroup$
    The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square.
    $endgroup$
    – Craig Hicks
    Mar 23 at 8:18










  • $begingroup$
    Thanks, I rolled back to my first answer.
    $endgroup$
    – farruhota
    Mar 23 at 8:29










  • $begingroup$
    But the graph was great! Bring back the graph!
    $endgroup$
    – Craig Hicks
    Mar 23 at 8:32










  • $begingroup$
    my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will.
    $endgroup$
    – farruhota
    Mar 23 at 8:35


















12












$begingroup$

First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.



You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.



So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.






share|cite|improve this answer









$endgroup$




















    6












    $begingroup$

    Let $A_e$ ($A$ early) be the event that bus $A$ arrives before $3$pm.



    Let $B_ell$ ($B$ late) be the event that bus $B$ arrives after $4$pm.



    Let $C$ be the union : $C=A_e cup B_ell$. Hence $$P(C)=P(A_e)+P(B_ell)-P(A_e cap B_ell)=P(A_e)+P(B_ell)-P(A_e)P(B_ell)$$



    Let $X$ be the event of interest ( bus $A$ arrives before bus $B$).



    What we know (don't we?) is that $P(X | C)=1$ and $P(X | overlineC)=0.5$



    Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$



    Can you go on from here ?






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
      $endgroup$
      – IrinaS
      Mar 23 at 5:26







    • 3




      $begingroup$
      @IrinaS - "A has already arrived before $B_ell$ happens" -- that would be $P(B_ell | A_ell)$. And you're right -- if $A_ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_ell$ and $B_ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A cup B)$ = 1
      $endgroup$
      – cag51
      Mar 23 at 7:23











    • $begingroup$
      @IrinaS What cag51 says above.
      $endgroup$
      – leonbloy
      Mar 23 at 14:01


















    1












    $begingroup$

    Define a new variable
    $Z = A-B = A+ (-B)$.



    Whenever $Z<0 Rightarrow A<B$ (A arrives before B)



    Since A and (-B) are independent, the pdf of Z is the convolution of the pdfs of A and (-B): $f_Z(z) = f_A(a)*f_-B(b)$.



    Solving the convolution graphically you get that:
    $$f_Z(z) = cases fracz+34, -3 leq z < -1
    \ frac1-z4, -1 leq z < 1$$



    Convolution between A and -B



    Now compute $P(Z<0)$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check.
      $endgroup$
      – Craig Hicks
      Mar 23 at 22:37


















    1












    $begingroup$

    $$int_s=2^4 int_t=3^5 p(a=s) p(b=t) delta(s<t) $$



    is the 2-d continuous integral equation. $delta(s<t)$ is 1 when $s<t$ and 0 otherwise. $delta(s<t)$ bisects the total area into two parts, the sum of which is 1.



    It is easily visualized and solved with a 2 x 2 hours block of time with a diagonal line $delta(s<t)$ cutting through one corner.



    The graph in farruhota's answer shows it.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      I wouldn't say "bisects". In other contexts, bisection implies equal parts.
      $endgroup$
      – David K
      Mar 23 at 14:40










    • $begingroup$
      @DavidK - I agree. "partitions" is a much better word choice.
      $endgroup$
      – Craig Hicks
      Mar 23 at 22:06


















    1












    $begingroup$

    What you missed in your approach to the question is that there is nothing in the problem statement that prevents bus $B$ from arriving after $4$ pm, and in fact (assuming uniform, independent distributions of the arrival times) half the time bus $B$ will arrive after $4$ pm,
    and in that case bus $A$ will have arrived first.



    Likewise, there is nothing that requires bus $A$ to arrive after $3$ pm so that bus $B$ has a chance to arrive first. Bus $A$ can just as likely arrive before $3$ pm.



    Let's take a frequentist approach. Suppose that these two buses run on this random schedule seven days a week, every day of the year. Let's watch them arrive for a few days and see what happens. Here's one possible way this might unfold:



    • On Monday, bus $A$ arrived at $2:38$ and bus $B$ arrived at $3:02$.
      Although bus $B$ arrived almost as quickly as it possibly can, bus $A$ still arrived first.


    • On Tuesday, bus $A$ arrived at $3:05$ and bus $B$ arrived at $3:42$.
      Bus $A$ arrived first.


    • On Wednesday, bus $A$ arrived at $2:50$ and bus $B$ arrived at $4:11$.
      Bus $A$ arrived first.


    • On Thursday, bus $A$ arrived at $3:57$ and bus $B$ arrived at $4:30$.
      Bus $A$ arrived first even though it was almost as late as it can be.


    • On Friday, bus $A$ arrived at $2:05$ and bus $B$ arrived at $4:56$.
      Bus $A$ arrived first.


    • On Saturday, bus $A$ arrived at $3:19$ and bus $B$ arrived at $3:17$.
      Finally we have observed an event in which bus $B$ arrived first!


    In the long run, if we keep track of the relative frequency of days like Monday
    (when bus $A$ arrives before $3$ pm and bus $B$ arrives between $3$ and $4$ pm),
    we'll find that the relative frequency approaches $1/4$ of all the days.
    To put it simply, in the long run $1/4$ of the days will be like Monday.



    Similarly, in the long run $1/4$ of the days will be like Wednesday and Friday, when bus $A$ arrived before $3$ pm and bus $B$ arrived after $4$ pm.
    Another $1/4$ of the days will be like Thursday, when bus $A$ arrived between $3$ and $4$ pm but bus $B$ arrives after $4$ pm.



    That leaves just $1/4$ of the days in the long run when bus $A$ and bus $B$ both arrive between $3$ and $4$ pm. One half of those days ($1/8$ of all days in the long run) will be like Tuesday, when bus $A$ arrived first, and the other half of those days ($1/8$ of all days in the long run) will be like Saturday, when bus $B$ arrived first.



    And that accounts for all possibilities. In one case, which happens $1/8$ of the time, bus $B$ arrives first. In all other cases bus $A$ arrives first.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Agree with your logic. This is a very good and simple explanation. Thank you, @David K.
      $endgroup$
      – IrinaS
      Mar 23 at 18:30


















    0












    $begingroup$

    The joint distribution for arrival times of A and B is
    $$P(A,B)d!Ad!B = frac14d!Ad!Bqquad 2<A<4, mathrmand ,3<B<5$$



    We need the probability that A is less than B
    $$P(A<B) = int_3^5 d!B int_2^min(B,4)d!A = frac78$$






    share|cite|improve this answer









    $endgroup$













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      7 Answers
      7






      active

      oldest

      votes








      7 Answers
      7






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      14












      $begingroup$

      Guide:



      1) Draw rectangle $2le xle 4,$ $3le yle 5$.



      2) The area of the rectangle is $4$, so pdf is $1/4$.



      3) Draw line $y=x$.



      4) Find area of the rectangle above the line, which is $7/2$.



      5) Finally, the required probability is $7/2cdot 1/4=7/8$.



      Here is the graph:



      $hspace2cm$ enter image description here



      Bus $A$ arriving before bus $B$:
      $$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$
      Bus $A$ arriving after bus $B$:
      $$E(3.7,3.3).$$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
        $endgroup$
        – IrinaS
        Mar 23 at 5:15










      • $begingroup$
        The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square.
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:18










      • $begingroup$
        Thanks, I rolled back to my first answer.
        $endgroup$
        – farruhota
        Mar 23 at 8:29










      • $begingroup$
        But the graph was great! Bring back the graph!
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:32










      • $begingroup$
        my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will.
        $endgroup$
        – farruhota
        Mar 23 at 8:35















      14












      $begingroup$

      Guide:



      1) Draw rectangle $2le xle 4,$ $3le yle 5$.



      2) The area of the rectangle is $4$, so pdf is $1/4$.



      3) Draw line $y=x$.



      4) Find area of the rectangle above the line, which is $7/2$.



      5) Finally, the required probability is $7/2cdot 1/4=7/8$.



      Here is the graph:



      $hspace2cm$ enter image description here



      Bus $A$ arriving before bus $B$:
      $$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$
      Bus $A$ arriving after bus $B$:
      $$E(3.7,3.3).$$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
        $endgroup$
        – IrinaS
        Mar 23 at 5:15










      • $begingroup$
        The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square.
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:18










      • $begingroup$
        Thanks, I rolled back to my first answer.
        $endgroup$
        – farruhota
        Mar 23 at 8:29










      • $begingroup$
        But the graph was great! Bring back the graph!
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:32










      • $begingroup$
        my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will.
        $endgroup$
        – farruhota
        Mar 23 at 8:35













      14












      14








      14





      $begingroup$

      Guide:



      1) Draw rectangle $2le xle 4,$ $3le yle 5$.



      2) The area of the rectangle is $4$, so pdf is $1/4$.



      3) Draw line $y=x$.



      4) Find area of the rectangle above the line, which is $7/2$.



      5) Finally, the required probability is $7/2cdot 1/4=7/8$.



      Here is the graph:



      $hspace2cm$ enter image description here



      Bus $A$ arriving before bus $B$:
      $$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$
      Bus $A$ arriving after bus $B$:
      $$E(3.7,3.3).$$






      share|cite|improve this answer











      $endgroup$



      Guide:



      1) Draw rectangle $2le xle 4,$ $3le yle 5$.



      2) The area of the rectangle is $4$, so pdf is $1/4$.



      3) Draw line $y=x$.



      4) Find area of the rectangle above the line, which is $7/2$.



      5) Finally, the required probability is $7/2cdot 1/4=7/8$.



      Here is the graph:



      $hspace2cm$ enter image description here



      Bus $A$ arriving before bus $B$:
      $$A(2.5,4.5),B(3.5,4.5),C(2.5,3.5),D(3.4,3.7).$$
      Bus $A$ arriving after bus $B$:
      $$E(3.7,3.3).$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 23 at 15:03









      David K

      55.6k345121




      55.6k345121










      answered Mar 23 at 4:24









      farruhotafarruhota

      21.9k2842




      21.9k2842











      • $begingroup$
        do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
        $endgroup$
        – IrinaS
        Mar 23 at 5:15










      • $begingroup$
        The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square.
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:18










      • $begingroup$
        Thanks, I rolled back to my first answer.
        $endgroup$
        – farruhota
        Mar 23 at 8:29










      • $begingroup$
        But the graph was great! Bring back the graph!
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:32










      • $begingroup$
        my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will.
        $endgroup$
        – farruhota
        Mar 23 at 8:35
















      • $begingroup$
        do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
        $endgroup$
        – IrinaS
        Mar 23 at 5:15










      • $begingroup$
        The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square.
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:18










      • $begingroup$
        Thanks, I rolled back to my first answer.
        $endgroup$
        – farruhota
        Mar 23 at 8:29










      • $begingroup$
        But the graph was great! Bring back the graph!
        $endgroup$
        – Craig Hicks
        Mar 23 at 8:32










      • $begingroup$
        my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will.
        $endgroup$
        – farruhota
        Mar 23 at 8:35















      $begingroup$
      do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
      $endgroup$
      – IrinaS
      Mar 23 at 5:15




      $begingroup$
      do we need the area above the x=y line, or below the line? I believe - under the line. Can you please clarify?
      $endgroup$
      – IrinaS
      Mar 23 at 5:15












      $begingroup$
      The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square.
      $endgroup$
      – Craig Hicks
      Mar 23 at 8:18




      $begingroup$
      The graph is a good idea but you need to get rid of the block F/G. Obviously the intersection is 2 hrs by 2 hrs - a square.
      $endgroup$
      – Craig Hicks
      Mar 23 at 8:18












      $begingroup$
      Thanks, I rolled back to my first answer.
      $endgroup$
      – farruhota
      Mar 23 at 8:29




      $begingroup$
      Thanks, I rolled back to my first answer.
      $endgroup$
      – farruhota
      Mar 23 at 8:29












      $begingroup$
      But the graph was great! Bring back the graph!
      $endgroup$
      – Craig Hicks
      Mar 23 at 8:32




      $begingroup$
      But the graph was great! Bring back the graph!
      $endgroup$
      – Craig Hicks
      Mar 23 at 8:32












      $begingroup$
      my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will.
      $endgroup$
      – farruhota
      Mar 23 at 8:35




      $begingroup$
      my first correct answer was from the phone, then I spoiled by drawing the graph on computer! Just kidding, I will.
      $endgroup$
      – farruhota
      Mar 23 at 8:35











      12












      $begingroup$

      First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.



      You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.



      So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.






      share|cite|improve this answer









      $endgroup$

















        12












        $begingroup$

        First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.



        You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.



        So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.






        share|cite|improve this answer









        $endgroup$















          12












          12








          12





          $begingroup$

          First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.



          You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.



          So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.






          share|cite|improve this answer









          $endgroup$



          First, I’ll assume that the probability distribution of each bus’s arrival time is uniform in its range and independent. I think that’s implicit in the question but you don’t actually say so.



          You’re mistaken when you say that if you know Bus A is arriving between $3$ and $4$, then there is an equal probability of either bus arriving first. There is a $50$% probability that Bus B arrives after $4$. The probabilities are equal only if you know that Bus A arrives between $3$ and $4$ and also that Bus B arrives between $3$ and $4$. That parlay occurs only $25$% of the time.



          So $75$% of the time, you know that Bus A arrives first, and Bus A still arrives first half of the remaining $25$% of the time. Thus, the probability that Bus A arrives first is $87.5$%.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 23 at 4:27









          Robert ShoreRobert Shore

          3,732324




          3,732324





















              6












              $begingroup$

              Let $A_e$ ($A$ early) be the event that bus $A$ arrives before $3$pm.



              Let $B_ell$ ($B$ late) be the event that bus $B$ arrives after $4$pm.



              Let $C$ be the union : $C=A_e cup B_ell$. Hence $$P(C)=P(A_e)+P(B_ell)-P(A_e cap B_ell)=P(A_e)+P(B_ell)-P(A_e)P(B_ell)$$



              Let $X$ be the event of interest ( bus $A$ arrives before bus $B$).



              What we know (don't we?) is that $P(X | C)=1$ and $P(X | overlineC)=0.5$



              Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$



              Can you go on from here ?






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
                $endgroup$
                – IrinaS
                Mar 23 at 5:26







              • 3




                $begingroup$
                @IrinaS - "A has already arrived before $B_ell$ happens" -- that would be $P(B_ell | A_ell)$. And you're right -- if $A_ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_ell$ and $B_ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A cup B)$ = 1
                $endgroup$
                – cag51
                Mar 23 at 7:23











              • $begingroup$
                @IrinaS What cag51 says above.
                $endgroup$
                – leonbloy
                Mar 23 at 14:01















              6












              $begingroup$

              Let $A_e$ ($A$ early) be the event that bus $A$ arrives before $3$pm.



              Let $B_ell$ ($B$ late) be the event that bus $B$ arrives after $4$pm.



              Let $C$ be the union : $C=A_e cup B_ell$. Hence $$P(C)=P(A_e)+P(B_ell)-P(A_e cap B_ell)=P(A_e)+P(B_ell)-P(A_e)P(B_ell)$$



              Let $X$ be the event of interest ( bus $A$ arrives before bus $B$).



              What we know (don't we?) is that $P(X | C)=1$ and $P(X | overlineC)=0.5$



              Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$



              Can you go on from here ?






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
                $endgroup$
                – IrinaS
                Mar 23 at 5:26







              • 3




                $begingroup$
                @IrinaS - "A has already arrived before $B_ell$ happens" -- that would be $P(B_ell | A_ell)$. And you're right -- if $A_ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_ell$ and $B_ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A cup B)$ = 1
                $endgroup$
                – cag51
                Mar 23 at 7:23











              • $begingroup$
                @IrinaS What cag51 says above.
                $endgroup$
                – leonbloy
                Mar 23 at 14:01













              6












              6








              6





              $begingroup$

              Let $A_e$ ($A$ early) be the event that bus $A$ arrives before $3$pm.



              Let $B_ell$ ($B$ late) be the event that bus $B$ arrives after $4$pm.



              Let $C$ be the union : $C=A_e cup B_ell$. Hence $$P(C)=P(A_e)+P(B_ell)-P(A_e cap B_ell)=P(A_e)+P(B_ell)-P(A_e)P(B_ell)$$



              Let $X$ be the event of interest ( bus $A$ arrives before bus $B$).



              What we know (don't we?) is that $P(X | C)=1$ and $P(X | overlineC)=0.5$



              Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$



              Can you go on from here ?






              share|cite|improve this answer











              $endgroup$



              Let $A_e$ ($A$ early) be the event that bus $A$ arrives before $3$pm.



              Let $B_ell$ ($B$ late) be the event that bus $B$ arrives after $4$pm.



              Let $C$ be the union : $C=A_e cup B_ell$. Hence $$P(C)=P(A_e)+P(B_ell)-P(A_e cap B_ell)=P(A_e)+P(B_ell)-P(A_e)P(B_ell)$$



              Let $X$ be the event of interest ( bus $A$ arrives before bus $B$).



              What we know (don't we?) is that $P(X | C)=1$ and $P(X | overlineC)=0.5$



              Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverlineC)=P(X | C) P(C) + P(X mid overlineC)P(overlineC)$$



              Can you go on from here ?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Mar 23 at 13:52

























              answered Mar 23 at 4:21









              leonbloyleonbloy

              42.2k647108




              42.2k647108











              • $begingroup$
                Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
                $endgroup$
                – IrinaS
                Mar 23 at 5:26







              • 3




                $begingroup$
                @IrinaS - "A has already arrived before $B_ell$ happens" -- that would be $P(B_ell | A_ell)$. And you're right -- if $A_ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_ell$ and $B_ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A cup B)$ = 1
                $endgroup$
                – cag51
                Mar 23 at 7:23











              • $begingroup$
                @IrinaS What cag51 says above.
                $endgroup$
                – leonbloy
                Mar 23 at 14:01
















              • $begingroup$
                Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
                $endgroup$
                – IrinaS
                Mar 23 at 5:26







              • 3




                $begingroup$
                @IrinaS - "A has already arrived before $B_ell$ happens" -- that would be $P(B_ell | A_ell)$. And you're right -- if $A_ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_ell$ and $B_ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A cup B)$ = 1
                $endgroup$
                – cag51
                Mar 23 at 7:23











              • $begingroup$
                @IrinaS What cag51 says above.
                $endgroup$
                – leonbloy
                Mar 23 at 14:01















              $begingroup$
              Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
              $endgroup$
              – IrinaS
              Mar 23 at 5:26





              $begingroup$
              Sorry for the question. I am confused with 𝑃(𝑋|𝐶)=1. Here is why - A has already arrived before 4 pm, before 𝐵ℓ happens, right? Yes, B can arrive after 4 pm, but it does not make any difference for the question we are asking which is - the probability of A arriving before B, before 4 pm, because A arrive between 2 and 4.
              $endgroup$
              – IrinaS
              Mar 23 at 5:26





              3




              3




              $begingroup$
              @IrinaS - "A has already arrived before $B_ell$ happens" -- that would be $P(B_ell | A_ell)$. And you're right -- if $A_ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_ell$ and $B_ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A cup B)$ = 1
              $endgroup$
              – cag51
              Mar 23 at 7:23





              $begingroup$
              @IrinaS - "A has already arrived before $B_ell$ happens" -- that would be $P(B_ell | A_ell)$. And you're right -- if $A_ell$ happens, then A arrives before B no matter what. But leonbloy is talking about the union of $A_ell$ and $B_ell$ -- A arrives before 3, OR B arrives after 4, OR both. In any of these cases, A arrives before B for sure, hence $P(X | A cup B)$ = 1
              $endgroup$
              – cag51
              Mar 23 at 7:23













              $begingroup$
              @IrinaS What cag51 says above.
              $endgroup$
              – leonbloy
              Mar 23 at 14:01




              $begingroup$
              @IrinaS What cag51 says above.
              $endgroup$
              – leonbloy
              Mar 23 at 14:01











              1












              $begingroup$

              Define a new variable
              $Z = A-B = A+ (-B)$.



              Whenever $Z<0 Rightarrow A<B$ (A arrives before B)



              Since A and (-B) are independent, the pdf of Z is the convolution of the pdfs of A and (-B): $f_Z(z) = f_A(a)*f_-B(b)$.



              Solving the convolution graphically you get that:
              $$f_Z(z) = cases fracz+34, -3 leq z < -1
              \ frac1-z4, -1 leq z < 1$$



              Convolution between A and -B



              Now compute $P(Z<0)$






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:37















              1












              $begingroup$

              Define a new variable
              $Z = A-B = A+ (-B)$.



              Whenever $Z<0 Rightarrow A<B$ (A arrives before B)



              Since A and (-B) are independent, the pdf of Z is the convolution of the pdfs of A and (-B): $f_Z(z) = f_A(a)*f_-B(b)$.



              Solving the convolution graphically you get that:
              $$f_Z(z) = cases fracz+34, -3 leq z < -1
              \ frac1-z4, -1 leq z < 1$$



              Convolution between A and -B



              Now compute $P(Z<0)$






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:37













              1












              1








              1





              $begingroup$

              Define a new variable
              $Z = A-B = A+ (-B)$.



              Whenever $Z<0 Rightarrow A<B$ (A arrives before B)



              Since A and (-B) are independent, the pdf of Z is the convolution of the pdfs of A and (-B): $f_Z(z) = f_A(a)*f_-B(b)$.



              Solving the convolution graphically you get that:
              $$f_Z(z) = cases fracz+34, -3 leq z < -1
              \ frac1-z4, -1 leq z < 1$$



              Convolution between A and -B



              Now compute $P(Z<0)$






              share|cite|improve this answer









              $endgroup$



              Define a new variable
              $Z = A-B = A+ (-B)$.



              Whenever $Z<0 Rightarrow A<B$ (A arrives before B)



              Since A and (-B) are independent, the pdf of Z is the convolution of the pdfs of A and (-B): $f_Z(z) = f_A(a)*f_-B(b)$.



              Solving the convolution graphically you get that:
              $$f_Z(z) = cases fracz+34, -3 leq z < -1
              \ frac1-z4, -1 leq z < 1$$



              Convolution between A and -B



              Now compute $P(Z<0)$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 23 at 8:40









              mazmaz

              1112




              1112











              • $begingroup$
                What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:37
















              • $begingroup$
                What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:37















              $begingroup$
              What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check.
              $endgroup$
              – Craig Hicks
              Mar 23 at 22:37




              $begingroup$
              What I like about this answer - (1) short description to get to computeable equation (2) the graph shows immediately the average, maximum, and minimum times that A arrives before B: 1 hr, 3hr, -1hr respectively, so the result is easy to check.
              $endgroup$
              – Craig Hicks
              Mar 23 at 22:37











              1












              $begingroup$

              $$int_s=2^4 int_t=3^5 p(a=s) p(b=t) delta(s<t) $$



              is the 2-d continuous integral equation. $delta(s<t)$ is 1 when $s<t$ and 0 otherwise. $delta(s<t)$ bisects the total area into two parts, the sum of which is 1.



              It is easily visualized and solved with a 2 x 2 hours block of time with a diagonal line $delta(s<t)$ cutting through one corner.



              The graph in farruhota's answer shows it.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                I wouldn't say "bisects". In other contexts, bisection implies equal parts.
                $endgroup$
                – David K
                Mar 23 at 14:40










              • $begingroup$
                @DavidK - I agree. "partitions" is a much better word choice.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:06















              1












              $begingroup$

              $$int_s=2^4 int_t=3^5 p(a=s) p(b=t) delta(s<t) $$



              is the 2-d continuous integral equation. $delta(s<t)$ is 1 when $s<t$ and 0 otherwise. $delta(s<t)$ bisects the total area into two parts, the sum of which is 1.



              It is easily visualized and solved with a 2 x 2 hours block of time with a diagonal line $delta(s<t)$ cutting through one corner.



              The graph in farruhota's answer shows it.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                I wouldn't say "bisects". In other contexts, bisection implies equal parts.
                $endgroup$
                – David K
                Mar 23 at 14:40










              • $begingroup$
                @DavidK - I agree. "partitions" is a much better word choice.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:06













              1












              1








              1





              $begingroup$

              $$int_s=2^4 int_t=3^5 p(a=s) p(b=t) delta(s<t) $$



              is the 2-d continuous integral equation. $delta(s<t)$ is 1 when $s<t$ and 0 otherwise. $delta(s<t)$ bisects the total area into two parts, the sum of which is 1.



              It is easily visualized and solved with a 2 x 2 hours block of time with a diagonal line $delta(s<t)$ cutting through one corner.



              The graph in farruhota's answer shows it.






              share|cite|improve this answer









              $endgroup$



              $$int_s=2^4 int_t=3^5 p(a=s) p(b=t) delta(s<t) $$



              is the 2-d continuous integral equation. $delta(s<t)$ is 1 when $s<t$ and 0 otherwise. $delta(s<t)$ bisects the total area into two parts, the sum of which is 1.



              It is easily visualized and solved with a 2 x 2 hours block of time with a diagonal line $delta(s<t)$ cutting through one corner.



              The graph in farruhota's answer shows it.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 23 at 8:45









              Craig HicksCraig Hicks

              1687




              1687











              • $begingroup$
                I wouldn't say "bisects". In other contexts, bisection implies equal parts.
                $endgroup$
                – David K
                Mar 23 at 14:40










              • $begingroup$
                @DavidK - I agree. "partitions" is a much better word choice.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:06
















              • $begingroup$
                I wouldn't say "bisects". In other contexts, bisection implies equal parts.
                $endgroup$
                – David K
                Mar 23 at 14:40










              • $begingroup$
                @DavidK - I agree. "partitions" is a much better word choice.
                $endgroup$
                – Craig Hicks
                Mar 23 at 22:06















              $begingroup$
              I wouldn't say "bisects". In other contexts, bisection implies equal parts.
              $endgroup$
              – David K
              Mar 23 at 14:40




              $begingroup$
              I wouldn't say "bisects". In other contexts, bisection implies equal parts.
              $endgroup$
              – David K
              Mar 23 at 14:40












              $begingroup$
              @DavidK - I agree. "partitions" is a much better word choice.
              $endgroup$
              – Craig Hicks
              Mar 23 at 22:06




              $begingroup$
              @DavidK - I agree. "partitions" is a much better word choice.
              $endgroup$
              – Craig Hicks
              Mar 23 at 22:06











              1












              $begingroup$

              What you missed in your approach to the question is that there is nothing in the problem statement that prevents bus $B$ from arriving after $4$ pm, and in fact (assuming uniform, independent distributions of the arrival times) half the time bus $B$ will arrive after $4$ pm,
              and in that case bus $A$ will have arrived first.



              Likewise, there is nothing that requires bus $A$ to arrive after $3$ pm so that bus $B$ has a chance to arrive first. Bus $A$ can just as likely arrive before $3$ pm.



              Let's take a frequentist approach. Suppose that these two buses run on this random schedule seven days a week, every day of the year. Let's watch them arrive for a few days and see what happens. Here's one possible way this might unfold:



              • On Monday, bus $A$ arrived at $2:38$ and bus $B$ arrived at $3:02$.
                Although bus $B$ arrived almost as quickly as it possibly can, bus $A$ still arrived first.


              • On Tuesday, bus $A$ arrived at $3:05$ and bus $B$ arrived at $3:42$.
                Bus $A$ arrived first.


              • On Wednesday, bus $A$ arrived at $2:50$ and bus $B$ arrived at $4:11$.
                Bus $A$ arrived first.


              • On Thursday, bus $A$ arrived at $3:57$ and bus $B$ arrived at $4:30$.
                Bus $A$ arrived first even though it was almost as late as it can be.


              • On Friday, bus $A$ arrived at $2:05$ and bus $B$ arrived at $4:56$.
                Bus $A$ arrived first.


              • On Saturday, bus $A$ arrived at $3:19$ and bus $B$ arrived at $3:17$.
                Finally we have observed an event in which bus $B$ arrived first!


              In the long run, if we keep track of the relative frequency of days like Monday
              (when bus $A$ arrives before $3$ pm and bus $B$ arrives between $3$ and $4$ pm),
              we'll find that the relative frequency approaches $1/4$ of all the days.
              To put it simply, in the long run $1/4$ of the days will be like Monday.



              Similarly, in the long run $1/4$ of the days will be like Wednesday and Friday, when bus $A$ arrived before $3$ pm and bus $B$ arrived after $4$ pm.
              Another $1/4$ of the days will be like Thursday, when bus $A$ arrived between $3$ and $4$ pm but bus $B$ arrives after $4$ pm.



              That leaves just $1/4$ of the days in the long run when bus $A$ and bus $B$ both arrive between $3$ and $4$ pm. One half of those days ($1/8$ of all days in the long run) will be like Tuesday, when bus $A$ arrived first, and the other half of those days ($1/8$ of all days in the long run) will be like Saturday, when bus $B$ arrived first.



              And that accounts for all possibilities. In one case, which happens $1/8$ of the time, bus $B$ arrives first. In all other cases bus $A$ arrives first.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                Agree with your logic. This is a very good and simple explanation. Thank you, @David K.
                $endgroup$
                – IrinaS
                Mar 23 at 18:30















              1












              $begingroup$

              What you missed in your approach to the question is that there is nothing in the problem statement that prevents bus $B$ from arriving after $4$ pm, and in fact (assuming uniform, independent distributions of the arrival times) half the time bus $B$ will arrive after $4$ pm,
              and in that case bus $A$ will have arrived first.



              Likewise, there is nothing that requires bus $A$ to arrive after $3$ pm so that bus $B$ has a chance to arrive first. Bus $A$ can just as likely arrive before $3$ pm.



              Let's take a frequentist approach. Suppose that these two buses run on this random schedule seven days a week, every day of the year. Let's watch them arrive for a few days and see what happens. Here's one possible way this might unfold:



              • On Monday, bus $A$ arrived at $2:38$ and bus $B$ arrived at $3:02$.
                Although bus $B$ arrived almost as quickly as it possibly can, bus $A$ still arrived first.


              • On Tuesday, bus $A$ arrived at $3:05$ and bus $B$ arrived at $3:42$.
                Bus $A$ arrived first.


              • On Wednesday, bus $A$ arrived at $2:50$ and bus $B$ arrived at $4:11$.
                Bus $A$ arrived first.


              • On Thursday, bus $A$ arrived at $3:57$ and bus $B$ arrived at $4:30$.
                Bus $A$ arrived first even though it was almost as late as it can be.


              • On Friday, bus $A$ arrived at $2:05$ and bus $B$ arrived at $4:56$.
                Bus $A$ arrived first.


              • On Saturday, bus $A$ arrived at $3:19$ and bus $B$ arrived at $3:17$.
                Finally we have observed an event in which bus $B$ arrived first!


              In the long run, if we keep track of the relative frequency of days like Monday
              (when bus $A$ arrives before $3$ pm and bus $B$ arrives between $3$ and $4$ pm),
              we'll find that the relative frequency approaches $1/4$ of all the days.
              To put it simply, in the long run $1/4$ of the days will be like Monday.



              Similarly, in the long run $1/4$ of the days will be like Wednesday and Friday, when bus $A$ arrived before $3$ pm and bus $B$ arrived after $4$ pm.
              Another $1/4$ of the days will be like Thursday, when bus $A$ arrived between $3$ and $4$ pm but bus $B$ arrives after $4$ pm.



              That leaves just $1/4$ of the days in the long run when bus $A$ and bus $B$ both arrive between $3$ and $4$ pm. One half of those days ($1/8$ of all days in the long run) will be like Tuesday, when bus $A$ arrived first, and the other half of those days ($1/8$ of all days in the long run) will be like Saturday, when bus $B$ arrived first.



              And that accounts for all possibilities. In one case, which happens $1/8$ of the time, bus $B$ arrives first. In all other cases bus $A$ arrives first.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                Agree with your logic. This is a very good and simple explanation. Thank you, @David K.
                $endgroup$
                – IrinaS
                Mar 23 at 18:30













              1












              1








              1





              $begingroup$

              What you missed in your approach to the question is that there is nothing in the problem statement that prevents bus $B$ from arriving after $4$ pm, and in fact (assuming uniform, independent distributions of the arrival times) half the time bus $B$ will arrive after $4$ pm,
              and in that case bus $A$ will have arrived first.



              Likewise, there is nothing that requires bus $A$ to arrive after $3$ pm so that bus $B$ has a chance to arrive first. Bus $A$ can just as likely arrive before $3$ pm.



              Let's take a frequentist approach. Suppose that these two buses run on this random schedule seven days a week, every day of the year. Let's watch them arrive for a few days and see what happens. Here's one possible way this might unfold:



              • On Monday, bus $A$ arrived at $2:38$ and bus $B$ arrived at $3:02$.
                Although bus $B$ arrived almost as quickly as it possibly can, bus $A$ still arrived first.


              • On Tuesday, bus $A$ arrived at $3:05$ and bus $B$ arrived at $3:42$.
                Bus $A$ arrived first.


              • On Wednesday, bus $A$ arrived at $2:50$ and bus $B$ arrived at $4:11$.
                Bus $A$ arrived first.


              • On Thursday, bus $A$ arrived at $3:57$ and bus $B$ arrived at $4:30$.
                Bus $A$ arrived first even though it was almost as late as it can be.


              • On Friday, bus $A$ arrived at $2:05$ and bus $B$ arrived at $4:56$.
                Bus $A$ arrived first.


              • On Saturday, bus $A$ arrived at $3:19$ and bus $B$ arrived at $3:17$.
                Finally we have observed an event in which bus $B$ arrived first!


              In the long run, if we keep track of the relative frequency of days like Monday
              (when bus $A$ arrives before $3$ pm and bus $B$ arrives between $3$ and $4$ pm),
              we'll find that the relative frequency approaches $1/4$ of all the days.
              To put it simply, in the long run $1/4$ of the days will be like Monday.



              Similarly, in the long run $1/4$ of the days will be like Wednesday and Friday, when bus $A$ arrived before $3$ pm and bus $B$ arrived after $4$ pm.
              Another $1/4$ of the days will be like Thursday, when bus $A$ arrived between $3$ and $4$ pm but bus $B$ arrives after $4$ pm.



              That leaves just $1/4$ of the days in the long run when bus $A$ and bus $B$ both arrive between $3$ and $4$ pm. One half of those days ($1/8$ of all days in the long run) will be like Tuesday, when bus $A$ arrived first, and the other half of those days ($1/8$ of all days in the long run) will be like Saturday, when bus $B$ arrived first.



              And that accounts for all possibilities. In one case, which happens $1/8$ of the time, bus $B$ arrives first. In all other cases bus $A$ arrives first.






              share|cite|improve this answer









              $endgroup$



              What you missed in your approach to the question is that there is nothing in the problem statement that prevents bus $B$ from arriving after $4$ pm, and in fact (assuming uniform, independent distributions of the arrival times) half the time bus $B$ will arrive after $4$ pm,
              and in that case bus $A$ will have arrived first.



              Likewise, there is nothing that requires bus $A$ to arrive after $3$ pm so that bus $B$ has a chance to arrive first. Bus $A$ can just as likely arrive before $3$ pm.



              Let's take a frequentist approach. Suppose that these two buses run on this random schedule seven days a week, every day of the year. Let's watch them arrive for a few days and see what happens. Here's one possible way this might unfold:



              • On Monday, bus $A$ arrived at $2:38$ and bus $B$ arrived at $3:02$.
                Although bus $B$ arrived almost as quickly as it possibly can, bus $A$ still arrived first.


              • On Tuesday, bus $A$ arrived at $3:05$ and bus $B$ arrived at $3:42$.
                Bus $A$ arrived first.


              • On Wednesday, bus $A$ arrived at $2:50$ and bus $B$ arrived at $4:11$.
                Bus $A$ arrived first.


              • On Thursday, bus $A$ arrived at $3:57$ and bus $B$ arrived at $4:30$.
                Bus $A$ arrived first even though it was almost as late as it can be.


              • On Friday, bus $A$ arrived at $2:05$ and bus $B$ arrived at $4:56$.
                Bus $A$ arrived first.


              • On Saturday, bus $A$ arrived at $3:19$ and bus $B$ arrived at $3:17$.
                Finally we have observed an event in which bus $B$ arrived first!


              In the long run, if we keep track of the relative frequency of days like Monday
              (when bus $A$ arrives before $3$ pm and bus $B$ arrives between $3$ and $4$ pm),
              we'll find that the relative frequency approaches $1/4$ of all the days.
              To put it simply, in the long run $1/4$ of the days will be like Monday.



              Similarly, in the long run $1/4$ of the days will be like Wednesday and Friday, when bus $A$ arrived before $3$ pm and bus $B$ arrived after $4$ pm.
              Another $1/4$ of the days will be like Thursday, when bus $A$ arrived between $3$ and $4$ pm but bus $B$ arrives after $4$ pm.



              That leaves just $1/4$ of the days in the long run when bus $A$ and bus $B$ both arrive between $3$ and $4$ pm. One half of those days ($1/8$ of all days in the long run) will be like Tuesday, when bus $A$ arrived first, and the other half of those days ($1/8$ of all days in the long run) will be like Saturday, when bus $B$ arrived first.



              And that accounts for all possibilities. In one case, which happens $1/8$ of the time, bus $B$ arrives first. In all other cases bus $A$ arrives first.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Mar 23 at 14:37









              David KDavid K

              55.6k345121




              55.6k345121











              • $begingroup$
                Agree with your logic. This is a very good and simple explanation. Thank you, @David K.
                $endgroup$
                – IrinaS
                Mar 23 at 18:30
















              • $begingroup$
                Agree with your logic. This is a very good and simple explanation. Thank you, @David K.
                $endgroup$
                – IrinaS
                Mar 23 at 18:30















              $begingroup$
              Agree with your logic. This is a very good and simple explanation. Thank you, @David K.
              $endgroup$
              – IrinaS
              Mar 23 at 18:30




              $begingroup$
              Agree with your logic. This is a very good and simple explanation. Thank you, @David K.
              $endgroup$
              – IrinaS
              Mar 23 at 18:30











              0












              $begingroup$

              The joint distribution for arrival times of A and B is
              $$P(A,B)d!Ad!B = frac14d!Ad!Bqquad 2<A<4, mathrmand ,3<B<5$$



              We need the probability that A is less than B
              $$P(A<B) = int_3^5 d!B int_2^min(B,4)d!A = frac78$$






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                The joint distribution for arrival times of A and B is
                $$P(A,B)d!Ad!B = frac14d!Ad!Bqquad 2<A<4, mathrmand ,3<B<5$$



                We need the probability that A is less than B
                $$P(A<B) = int_3^5 d!B int_2^min(B,4)d!A = frac78$$






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  The joint distribution for arrival times of A and B is
                  $$P(A,B)d!Ad!B = frac14d!Ad!Bqquad 2<A<4, mathrmand ,3<B<5$$



                  We need the probability that A is less than B
                  $$P(A<B) = int_3^5 d!B int_2^min(B,4)d!A = frac78$$






                  share|cite|improve this answer









                  $endgroup$



                  The joint distribution for arrival times of A and B is
                  $$P(A,B)d!Ad!B = frac14d!Ad!Bqquad 2<A<4, mathrmand ,3<B<5$$



                  We need the probability that A is less than B
                  $$P(A<B) = int_3^5 d!B int_2^min(B,4)d!A = frac78$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 23 at 10:12









                  user3856370user3856370

                  1514




                  1514



























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