The probability of Bus A arriving before Bus B
$begingroup$
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
asked 2 hours 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.
$endgroup$
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
asked 2 hours 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
58 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
asked 2 hours 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.
$endgroup$
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
asked 2 hours 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 2 hours 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.
edited 1 hour ago
IrinaS
asked 2 hours 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 2 hours ago
IrinaSIrinaS
62
asked 2 hours ago
IrinaSIrinaS
62
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.
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.
IrinaS is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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
58 mins ago
add a comment |
$begingroup$
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
58 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
58 mins ago
$begingroup$
Yes, the arrival of buses is independent events.
$endgroup$
– IrinaS
58 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 ?
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
17 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 $B$:
$$E(3.7,3.3).$$
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
28 mins ago
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 $87.5$%.
answered 1 hour ago
Robert ShoreRobert Shore
3,410323
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 ?
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
17 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 ?
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
17 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 ?
answered 1 hour ago
leonbloyleonbloy
41.8k647108
$endgroup$
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 ?
answered 1 hour ago
leonbloyleonbloy
41.8k647108
answered 1 hour ago
leonbloyleonbloy
41.8k647108
answered 1 hour ago
leonbloyleonbloy
41.8k647108
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
17 mins ago
add a comment |
$begingroup$
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
17 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
17 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
17 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 $B$:
$$E(3.7,3.3).$$
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
28 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 $B$:
$$E(3.7,3.3).$$
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
28 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 $B$:
$$E(3.7,3.3).$$
answered 1 hour ago
farruhotafarruhota
21.4k2841
$endgroup$
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:
$hspace{2cm}$
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 $B$:
$$E(3.7,3.3).$$
answered 1 hour ago
farruhotafarruhota
21.4k2841
edited 18 mins ago
answered 1 hour ago
farruhotafarruhota
21.4k2841
answered 1 hour ago
farruhotafarruhota
21.4k2841
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
28 mins ago
add a comment |
$begingroup$
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
28 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
28 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
28 mins ago
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 $87.5$%.
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 $87.5$%.
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 $87.5$%.
answered 1 hour ago
Robert ShoreRobert Shore
3,410323
$endgroup$
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 $87.5$%.
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
answered 1 hour ago
Robert ShoreRobert Shore
3,410323
3,410323
add a comment |
add a comment |
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
IrinaS is a new contributor. Be nice, and check out our Code of Conduct.
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Using conditional probability I think it must be. $P(A/B')$ proceeding from here gives us ans $1/22$ is it correct?
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– Vimath
1 hour ago
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Yes, the arrival of buses is independent events.
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– IrinaS
58 mins ago