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The probability of Bus A arriving before Bus B

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The probability of Bus A arriving before Bus B

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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
  39 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
  39 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

$endgroup$

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

share|cite|improve this answer

edited 45 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
  10 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 58 mins 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

$endgroup$

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

$endgroup$

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

share|cite|improve this answer

edited 45 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
  10 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 $4.5cdot 1/5=9/10$.

share|cite|improve this answer

edited 45 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
  10 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 $4.5cdot 1/5=9/10$.

share|cite|improve this answer

edited 45 mins ago

answered 1 hour ago

farruhotafarruhota

21.4k2841

$endgroup$

share|cite|improve this answer

edited 45 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 58 mins 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 58 mins 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 58 mins ago

Robert ShoreRobert Shore

3,410323

$endgroup$

share|cite|improve this answer

answered 58 mins ago

Robert ShoreRobert Shore

3,410323

answered 58 mins ago

Robert ShoreRobert Shore

3,410323

answered 58 mins ago

Robert ShoreRobert Shore

3,410323