Probability of bus arriving at given time
Am i calculating this correctly. The metro stops by every 8 minutes and the bus departs every 10 minutes. If i don't follow the schedules but arrive at the stop randomly. With what probability.
A) I have to wait for the metro more than 5 minutes.
$3/8 = 37.5%$
B) Both metro and bus leave after waiting less than 2 minutes.
$2/8 * 2/8 = 6.25%$
probability
asked Dec 27 '18 at 1:16
James Leaf
135
|
show 1 more comment
Am i calculating this correctly. The metro stops by every 8 minutes and the bus departs every 10 minutes. If i don't follow the schedules but arrive at the stop randomly. With what probability.
A) I have to wait for the metro more than 5 minutes.
$3/8 = 37.5%$
B) Both metro and bus leave after waiting less than 2 minutes.
$2/8 * 2/8 = 6.25%$
probability
asked Dec 27 '18 at 1:16
James Leaf
135
Shouldn't B be $2/8 cdot 2/10$?
– Math1000
Dec 27 '18 at 1:28
Probably, that's why i'm asking this question to see if i someone knew if i did it correctly. Someone more skilled in math than i.
– James Leaf
Dec 27 '18 at 1:40
"arrive at the stops randomly"— with what distribution?
– Alex
Dec 27 '18 at 1:54
What i mean by that is that i arrive at the stop at a random time which leads to the given questions.
– James Leaf
Dec 27 '18 at 1:58
As in, you're equally likely to arrive at the bus stop at any time?
– Alex
Dec 27 '18 at 1:59
|
show 1 more comment
Am i calculating this correctly. The metro stops by every 8 minutes and the bus departs every 10 minutes. If i don't follow the schedules but arrive at the stop randomly. With what probability.
A) I have to wait for the metro more than 5 minutes.
$3/8 = 37.5%$
B) Both metro and bus leave after waiting less than 2 minutes.
$2/8 * 2/8 = 6.25%$
probability
asked Dec 27 '18 at 1:16
James Leaf
135
Am i calculating this correctly. The metro stops by every 8 minutes and the bus departs every 10 minutes. If i don't follow the schedules but arrive at the stop randomly. With what probability.
A) I have to wait for the metro more than 5 minutes.
$3/8 = 37.5%$
B) Both metro and bus leave after waiting less than 2 minutes.
$2/8 * 2/8 = 6.25%$
probability
probability
asked Dec 27 '18 at 1:16
James Leaf
135
asked Dec 27 '18 at 1:16
James Leaf
135
edited Dec 27 '18 at 1:57
asked Dec 27 '18 at 1:16
James Leaf
135
asked Dec 27 '18 at 1:16
James Leaf
135
asked Dec 27 '18 at 1:16
James Leaf
135
135
Shouldn't B be $2/8 cdot 2/10$?
– Math1000
Dec 27 '18 at 1:28
Probably, that's why i'm asking this question to see if i someone knew if i did it correctly. Someone more skilled in math than i.
– James Leaf
Dec 27 '18 at 1:40
"arrive at the stops randomly"— with what distribution?
– Alex
Dec 27 '18 at 1:54
What i mean by that is that i arrive at the stop at a random time which leads to the given questions.
– James Leaf
Dec 27 '18 at 1:58
As in, you're equally likely to arrive at the bus stop at any time?
– Alex
Dec 27 '18 at 1:59
|
show 1 more comment
Shouldn't B be $2/8 cdot 2/10$?
– Math1000
Dec 27 '18 at 1:28
Probably, that's why i'm asking this question to see if i someone knew if i did it correctly. Someone more skilled in math than i.
– James Leaf
Dec 27 '18 at 1:40
"arrive at the stops randomly"— with what distribution?
– Alex
Dec 27 '18 at 1:54
What i mean by that is that i arrive at the stop at a random time which leads to the given questions.
– James Leaf
Dec 27 '18 at 1:58
As in, you're equally likely to arrive at the bus stop at any time?
– Alex
Dec 27 '18 at 1:59
Shouldn't B be $2/8 cdot 2/10$?
– Math1000
Dec 27 '18 at 1:28
Shouldn't B be $2/8 cdot 2/10$?
– Math1000
Dec 27 '18 at 1:28
Probably, that's why i'm asking this question to see if i someone knew if i did it correctly. Someone more skilled in math than i.
– James Leaf
Dec 27 '18 at 1:40
Probably, that's why i'm asking this question to see if i someone knew if i did it correctly. Someone more skilled in math than i.
– James Leaf
Dec 27 '18 at 1:40
"arrive at the stops randomly"— with what distribution?
– Alex
Dec 27 '18 at 1:54
"arrive at the stops randomly"— with what distribution?
– Alex
Dec 27 '18 at 1:54
What i mean by that is that i arrive at the stop at a random time which leads to the given questions.
– James Leaf
Dec 27 '18 at 1:58
What i mean by that is that i arrive at the stop at a random time which leads to the given questions.
– James Leaf
Dec 27 '18 at 1:58
As in, you're equally likely to arrive at the bus stop at any time?
– Alex
Dec 27 '18 at 1:59
As in, you're equally likely to arrive at the bus stop at any time?
– Alex
Dec 27 '18 at 1:59
|
show 1 more comment
1 Answer
1
active
oldest
votes
Let's assume that you will only arrive "on the minute" (e.g. at 10:05, never at 10:05 and 15 seconds).
A) When you arrive, there are $8$ possibilities: you arrive $0$ minutes before the metro leaves (i.e. when it is leaving; in this case I'll assume you can get on the metro), you arrive $1$ minute before the metro leaves, you arrive $2$ minutes before the metro leaves, ..., you arrive $7$ minutes before the metro leaves. You will have to wait more than $5$ minutes only if you arrive $6$ minutes before the metro leaves or $7$ minutes before it leaves. Since each possibility is equally likely, and $2$ of the $8$ possibilities require waiting more than $5$ minutes, the probability is $2/8$.
(Aside: your answer of $3/5$ would be the probability of waiting at least $5$ minutes)
B) This one is more complicated. Let's assume the first bus and metro arrive at the same time (say time $0$). Then a metro arrives at times $0, 8, 16, 24, 32, 40 = 0$. A bus arrives at times $0, 10, 20, 30, 40 = 0$. In both cases I treat $40$ as "wrapping around" to $0$ because that's when the full cycle of metro-bus arrivals repeats.
If both the metro and the bus leave after less than $2$ minutes of waiting, then they must leave within less than $2$ minutes of one another (for example, if they left with a $3$ minute gap between them you'd have to wait for at least $3$ minutes). The only time this happens is at minute $0=40$ in the cycle. If you want to wait for less than $2$ minutes, you'd need to arrive at either minute $39$ or minute $40$. So there's a $2/40$ probability.
If you're interested in the probability of waiting at most $2$ minutes (i.e. you're okay if you have to wait $2$ minutes, rather than less than $2$ minutes). Now you can achieve this by arriving at minute $8$, minute $30$, minute $38$, minute $39$, or minute $40$. Since each of these $5$ arrivals is equally likely, you have a probability of $5/40$.
answered Dec 27 '18 at 2:25
Alex
1777
add a comment |
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1 Answer
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Let's assume that you will only arrive "on the minute" (e.g. at 10:05, never at 10:05 and 15 seconds).
A) When you arrive, there are $8$ possibilities: you arrive $0$ minutes before the metro leaves (i.e. when it is leaving; in this case I'll assume you can get on the metro), you arrive $1$ minute before the metro leaves, you arrive $2$ minutes before the metro leaves, ..., you arrive $7$ minutes before the metro leaves. You will have to wait more than $5$ minutes only if you arrive $6$ minutes before the metro leaves or $7$ minutes before it leaves. Since each possibility is equally likely, and $2$ of the $8$ possibilities require waiting more than $5$ minutes, the probability is $2/8$.
(Aside: your answer of $3/5$ would be the probability of waiting at least $5$ minutes)
B) This one is more complicated. Let's assume the first bus and metro arrive at the same time (say time $0$). Then a metro arrives at times $0, 8, 16, 24, 32, 40 = 0$. A bus arrives at times $0, 10, 20, 30, 40 = 0$. In both cases I treat $40$ as "wrapping around" to $0$ because that's when the full cycle of metro-bus arrivals repeats.
If both the metro and the bus leave after less than $2$ minutes of waiting, then they must leave within less than $2$ minutes of one another (for example, if they left with a $3$ minute gap between them you'd have to wait for at least $3$ minutes). The only time this happens is at minute $0=40$ in the cycle. If you want to wait for less than $2$ minutes, you'd need to arrive at either minute $39$ or minute $40$. So there's a $2/40$ probability.
If you're interested in the probability of waiting at most $2$ minutes (i.e. you're okay if you have to wait $2$ minutes, rather than less than $2$ minutes). Now you can achieve this by arriving at minute $8$, minute $30$, minute $38$, minute $39$, or minute $40$. Since each of these $5$ arrivals is equally likely, you have a probability of $5/40$.
answered Dec 27 '18 at 2:25
Alex
1777
add a comment |
Let's assume that you will only arrive "on the minute" (e.g. at 10:05, never at 10:05 and 15 seconds).
A) When you arrive, there are $8$ possibilities: you arrive $0$ minutes before the metro leaves (i.e. when it is leaving; in this case I'll assume you can get on the metro), you arrive $1$ minute before the metro leaves, you arrive $2$ minutes before the metro leaves, ..., you arrive $7$ minutes before the metro leaves. You will have to wait more than $5$ minutes only if you arrive $6$ minutes before the metro leaves or $7$ minutes before it leaves. Since each possibility is equally likely, and $2$ of the $8$ possibilities require waiting more than $5$ minutes, the probability is $2/8$.
(Aside: your answer of $3/5$ would be the probability of waiting at least $5$ minutes)
B) This one is more complicated. Let's assume the first bus and metro arrive at the same time (say time $0$). Then a metro arrives at times $0, 8, 16, 24, 32, 40 = 0$. A bus arrives at times $0, 10, 20, 30, 40 = 0$. In both cases I treat $40$ as "wrapping around" to $0$ because that's when the full cycle of metro-bus arrivals repeats.
If both the metro and the bus leave after less than $2$ minutes of waiting, then they must leave within less than $2$ minutes of one another (for example, if they left with a $3$ minute gap between them you'd have to wait for at least $3$ minutes). The only time this happens is at minute $0=40$ in the cycle. If you want to wait for less than $2$ minutes, you'd need to arrive at either minute $39$ or minute $40$. So there's a $2/40$ probability.
If you're interested in the probability of waiting at most $2$ minutes (i.e. you're okay if you have to wait $2$ minutes, rather than less than $2$ minutes). Now you can achieve this by arriving at minute $8$, minute $30$, minute $38$, minute $39$, or minute $40$. Since each of these $5$ arrivals is equally likely, you have a probability of $5/40$.
answered Dec 27 '18 at 2:25
Alex
1777
add a comment |
Let's assume that you will only arrive "on the minute" (e.g. at 10:05, never at 10:05 and 15 seconds).
A) When you arrive, there are $8$ possibilities: you arrive $0$ minutes before the metro leaves (i.e. when it is leaving; in this case I'll assume you can get on the metro), you arrive $1$ minute before the metro leaves, you arrive $2$ minutes before the metro leaves, ..., you arrive $7$ minutes before the metro leaves. You will have to wait more than $5$ minutes only if you arrive $6$ minutes before the metro leaves or $7$ minutes before it leaves. Since each possibility is equally likely, and $2$ of the $8$ possibilities require waiting more than $5$ minutes, the probability is $2/8$.
(Aside: your answer of $3/5$ would be the probability of waiting at least $5$ minutes)
B) This one is more complicated. Let's assume the first bus and metro arrive at the same time (say time $0$). Then a metro arrives at times $0, 8, 16, 24, 32, 40 = 0$. A bus arrives at times $0, 10, 20, 30, 40 = 0$. In both cases I treat $40$ as "wrapping around" to $0$ because that's when the full cycle of metro-bus arrivals repeats.
If both the metro and the bus leave after less than $2$ minutes of waiting, then they must leave within less than $2$ minutes of one another (for example, if they left with a $3$ minute gap between them you'd have to wait for at least $3$ minutes). The only time this happens is at minute $0=40$ in the cycle. If you want to wait for less than $2$ minutes, you'd need to arrive at either minute $39$ or minute $40$. So there's a $2/40$ probability.
If you're interested in the probability of waiting at most $2$ minutes (i.e. you're okay if you have to wait $2$ minutes, rather than less than $2$ minutes). Now you can achieve this by arriving at minute $8$, minute $30$, minute $38$, minute $39$, or minute $40$. Since each of these $5$ arrivals is equally likely, you have a probability of $5/40$.
answered Dec 27 '18 at 2:25
Alex
1777
Let's assume that you will only arrive "on the minute" (e.g. at 10:05, never at 10:05 and 15 seconds).
A) When you arrive, there are $8$ possibilities: you arrive $0$ minutes before the metro leaves (i.e. when it is leaving; in this case I'll assume you can get on the metro), you arrive $1$ minute before the metro leaves, you arrive $2$ minutes before the metro leaves, ..., you arrive $7$ minutes before the metro leaves. You will have to wait more than $5$ minutes only if you arrive $6$ minutes before the metro leaves or $7$ minutes before it leaves. Since each possibility is equally likely, and $2$ of the $8$ possibilities require waiting more than $5$ minutes, the probability is $2/8$.
(Aside: your answer of $3/5$ would be the probability of waiting at least $5$ minutes)
B) This one is more complicated. Let's assume the first bus and metro arrive at the same time (say time $0$). Then a metro arrives at times $0, 8, 16, 24, 32, 40 = 0$. A bus arrives at times $0, 10, 20, 30, 40 = 0$. In both cases I treat $40$ as "wrapping around" to $0$ because that's when the full cycle of metro-bus arrivals repeats.
If both the metro and the bus leave after less than $2$ minutes of waiting, then they must leave within less than $2$ minutes of one another (for example, if they left with a $3$ minute gap between them you'd have to wait for at least $3$ minutes). The only time this happens is at minute $0=40$ in the cycle. If you want to wait for less than $2$ minutes, you'd need to arrive at either minute $39$ or minute $40$. So there's a $2/40$ probability.
If you're interested in the probability of waiting at most $2$ minutes (i.e. you're okay if you have to wait $2$ minutes, rather than less than $2$ minutes). Now you can achieve this by arriving at minute $8$, minute $30$, minute $38$, minute $39$, or minute $40$. Since each of these $5$ arrivals is equally likely, you have a probability of $5/40$.
answered Dec 27 '18 at 2:25
Alex
1777
answered Dec 27 '18 at 2:25
Alex
1777
answered Dec 27 '18 at 2:25
Alex
1777
answered Dec 27 '18 at 2:25
Alex
1777
1777
add a comment |
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Shouldn't B be $2/8 cdot 2/10$?
– Math1000
Dec 27 '18 at 1:28
Probably, that's why i'm asking this question to see if i someone knew if i did it correctly. Someone more skilled in math than i.
– James Leaf
Dec 27 '18 at 1:40
"arrive at the stops randomly"— with what distribution?
– Alex
Dec 27 '18 at 1:54
What i mean by that is that i arrive at the stop at a random time which leads to the given questions.
– James Leaf
Dec 27 '18 at 1:58
As in, you're equally likely to arrive at the bus stop at any time?
– Alex
Dec 27 '18 at 1:59