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?






probability







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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
    43 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
    43 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

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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
    43 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
    43 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
43 mins ago




$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
43 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









$endgroup$













  • $begingroup$
    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
    2 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 $4.5cdot 1/5=9/10$.



Here is the graph:



$hspace{2cm}$enter image description here



Bus $A$ arriving before bus $B$:
$$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
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
    13 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









$endgroup$













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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









    $endgroup$













    • $begingroup$
      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
      2 mins ago


















    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









    $endgroup$













    • $begingroup$
      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
      2 mins ago
















    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












    • $begingroup$
      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
      2 mins ago




















    • $begingroup$
      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
      2 mins ago


















    $begingroup$
    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
    2 mins ago






    $begingroup$
    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
    2 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 $4.5cdot 1/5=9/10$.



    Here is the graph:



    $hspace{2cm}$enter image description here



    Bus $A$ arriving before bus $B$:
    $$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
    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
      13 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 $4.5cdot 1/5=9/10$.



    Here is the graph:



    $hspace{2cm}$enter image description here



    Bus $A$ arriving before bus $B$:
    $$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
    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
      13 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 $4.5cdot 1/5=9/10$.



    Here is the graph:



    $hspace{2cm}$enter image description here



    Bus $A$ arriving before bus $B$:
    $$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
    Bus $A$ arriving after bus $B$:
    $$E(3.7,3.3).$$






    share|cite|improve this answer











    $endgroup$



    Guide:



    1) Draw rectangle $2le xle 4, 3le yle 5$, 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 $4.5cdot 1/5=9/10$.



    Here is the graph:



    $hspace{2cm}$enter image description here



    Bus $A$ arriving before bus $B$:
    $$A(2.5,4.5), B(3.5,4.5), C(2.5,3.5), D(3.4, 3.7), F(2.3,-), G(2.7,-).$$
    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 3 mins ago

























    answered 1 hour 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
      13 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
      13 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
    13 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
    13 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 1 hour ago









        Robert ShoreRobert Shore

        3,410323




        3,410323






















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            Fluorita

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            Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac...
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            Hulsita

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            Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà ...
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            Península de Txukotka

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            Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
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