The probability of Bus A arriving before Bus B












1












$begingroup$


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?






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  • $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    1 hour ago










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    38 mins ago
















1












$begingroup$


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?






probability







share|cite|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    1 hour ago










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    38 mins ago














1












1








1





$begingroup$


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?






probability







share|cite|improve this question









New contributor




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







$endgroup$




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




probability






share|cite|improve this question









New contributor




IrinaS 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 question









New contributor




IrinaS 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 question




share|cite|improve this question








edited 1 hour ago







IrinaS













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asked 1 hour ago









IrinaSIrinaS

62




62




New contributor




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





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






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Check out our Code of Conduct.












  • $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    1 hour ago










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    38 mins ago


















  • $begingroup$
    Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
    $endgroup$
    – Vimath
    1 hour ago










  • $begingroup$
    Yes, the arrival of buses is independent events.
    $endgroup$
    – IrinaS
    38 mins ago
















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




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












$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
38 mins ago




$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
38 mins ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

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



Let $B_ell$ be the event that $B$ arrives after $4$pm.



Let $C$ be the union : $C=A_e cup B_ell$.



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



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



Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverline{C})=P(X | C) P(C) + P(X mid overline{C})P(overline{C})$$



Can you go on from here ?






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Guide:



    1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.



    2) The total area of the rectangle and square is $5$, so pdf is $1/5$.



    3) Find total area of the square and rectangle above the line, which is $4.5$.



    5) Finally, the required probability is $4.5cdot 1/5=9/10$.






    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
      8 mins ago



















    0












    $begingroup$

    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









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

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



      Let $B_ell$ be the event that $B$ arrives after $4$pm.



      Let $C$ be the union : $C=A_e cup B_ell$.



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



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



      Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverline{C})=P(X | C) P(C) + P(X mid overline{C})P(overline{C})$$



      Can you go on from here ?






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

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



        Let $B_ell$ be the event that $B$ arrives after $4$pm.



        Let $C$ be the union : $C=A_e cup B_ell$.



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



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



        Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverline{C})=P(X | C) P(C) + P(X mid overline{C})P(overline{C})$$



        Can you go on from here ?






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

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



          Let $B_ell$ be the event that $B$ arrives after $4$pm.



          Let $C$ be the union : $C=A_e cup B_ell$.



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



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



          Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverline{C})=P(X | C) P(C) + P(X mid overline{C})P(overline{C})$$



          Can you go on from here ?






          share|cite|improve this answer









          $endgroup$



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



          Let $B_ell$ be the event that $B$ arrives after $4$pm.



          Let $C$ be the union : $C=A_e cup B_ell$.



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



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



          Then we can write (total probability) $$P(X) = P(X cap C) + P(X capoverline{C})=P(X | C) P(C) + P(X mid overline{C})P(overline{C})$$



          Can you go on from here ?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          leonbloyleonbloy

          41.8k647108




          41.8k647108























              1












              $begingroup$

              Guide:



              1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.



              2) The total area of the rectangle and square is $5$, so pdf is $1/5$.



              3) Find total area of the square and rectangle above the line, which is $4.5$.



              5) Finally, the required probability is $4.5cdot 1/5=9/10$.






              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
                8 mins ago
















              1












              $begingroup$

              Guide:



              1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.



              2) The total area of the rectangle and square is $5$, so pdf is $1/5$.



              3) Find total area of the square and rectangle above the line, which is $4.5$.



              5) Finally, the required probability is $4.5cdot 1/5=9/10$.






              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
                8 mins ago














              1












              1








              1





              $begingroup$

              Guide:



              1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.



              2) The total area of the rectangle and square is $5$, so pdf is $1/5$.



              3) Find total area of the square and rectangle above the line, which is $4.5$.



              5) Finally, the required probability is $4.5cdot 1/5=9/10$.






              share|cite|improve this answer











              $endgroup$



              Guide:



              1) Draw rectangle $2le xle 4, 3le yle 5$, square $2le xle 3, 2le yle 3$ and line $y=x$.



              2) The total area of the rectangle and square is $5$, so pdf is $1/5$.



              3) Find total area of the square and rectangle above the line, which is $4.5$.



              5) Finally, the required probability is $4.5cdot 1/5=9/10$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 43 mins ago

























              answered 59 mins ago









              farruhotafarruhota

              21.4k2841




              21.4k2841












              • $begingroup$
                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
                8 mins ago


















              • $begingroup$
                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
                8 mins ago
















              $begingroup$
              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
              8 mins ago




              $begingroup$
              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
              8 mins ago











              0












              $begingroup$

              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









              $endgroup$


















                0












                $begingroup$

                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









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  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 56 mins ago









                  Robert ShoreRobert Shore

                  3,410323




                  3,410323






















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