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The probability of Bus A arriving before Bus B

Expected 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

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

edited 1 hour ago

IrinaS

asked 1 hour ago

IrinaSIrinaS

62

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
  56 mins ago
  

  
  
  
  

add a comment | 

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

edited 1 hour ago

IrinaS

asked 1 hour ago

IrinaSIrinaS

62

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
  56 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 1 hour ago

IrinaS

asked 1 hour ago

IrinaSIrinaS

62

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.

$endgroup$

share|cite|improve this question

edited 1 hour ago

IrinaS

asked 1 hour ago

IrinaSIrinaS

62

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

edited 1 hour ago

IrinaS

asked 1 hour ago

IrinaSIrinaS

62

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.

asked 1 hour ago

IrinaSIrinaS

62

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.

asked 1 hour ago

IrinaSIrinaS

62

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

answered 1 hour ago

leonbloyleonbloy

41.8k647108

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  15 mins ago
  
  
  

  
  
  
  

add a comment | 

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:

$hspace{2cm}$

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

edited 16 mins ago

answered 1 hour ago

farruhotafarruhota

21.4k2841

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  26 mins ago
  

  
  
  
  

add a comment | 

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

answered 1 hour ago

Robert ShoreRobert Shore

3,410323

$endgroup$

add a comment | 

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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

answered 1 hour ago

leonbloyleonbloy

41.8k647108

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  15 mins ago
  
  
  

  
  
  
  

add a comment | 

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

answered 1 hour ago

leonbloyleonbloy

41.8k647108

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  15 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 1 hour ago

leonbloyleonbloy

41.8k647108

$endgroup$

share|cite|improve this answer

answered 1 hour ago

leonbloyleonbloy

41.8k647108

answered 1 hour ago

leonbloyleonbloy

41.8k647108

answered 1 hour ago

leonbloyleonbloy

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

Here is the graph:

$hspace{2cm}$

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

edited 16 mins ago

answered 1 hour ago

farruhotafarruhota

21.4k2841

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  26 mins ago
  

  
  
  
  

add a comment | 

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:

$hspace{2cm}$

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

edited 16 mins ago

answered 1 hour ago

farruhotafarruhota

21.4k2841

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  26 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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:

$hspace{2cm}$

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

edited 16 mins ago

answered 1 hour ago

farruhotafarruhota

21.4k2841

$endgroup$

share|cite|improve this answer

edited 16 mins ago

answered 1 hour ago

farruhotafarruhota

21.4k2841

answered 1 hour ago

farruhotafarruhota

21.4k2841

answered 1 hour ago

farruhotafarruhota

21.4k2841

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

answered 1 hour ago

Robert ShoreRobert Shore

3,410323

$endgroup$

add a comment | 

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

answered 1 hour ago

Robert ShoreRobert Shore

3,410323

$endgroup$

add a comment | 

$begingroup$

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

answered 1 hour ago

Robert ShoreRobert Shore

3,410323

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Robert ShoreRobert Shore

3,410323

answered 1 hour ago

Robert ShoreRobert Shore

3,410323

answered 1 hour ago

Robert ShoreRobert Shore

3,410323