Mixing
Interpretation of a seemingly simple probability question
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I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:
There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?
What is your answer, and why? Thanks in advance.
probability
asked Sep 6 at 18:22
JPMaverick
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I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:
There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?
What is your answer, and why? Thanks in advance.
probability
asked Sep 6 at 18:22
JPMaverick
311
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JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
6
down vote
favorite
up vote
6
down vote
favorite
I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:
There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?
What is your answer, and why? Thanks in advance.
probability
asked Sep 6 at 18:22
JPMaverick
311
New contributor
JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I was asked this question in an interview, and did not initially answer correctly though I still think my interpretation may have been the correct one. The question was:
There are two delivery trucks, A and B. A makes deliveries between 8am and 10am, and B makes deliveries between 9am and 11am. The deliveries are uniformly distributed for both. What is the probability that any given delivery from B will take place before any delivery from A?
What is your answer, and why? Thanks in advance.
probability
asked Sep 6 at 18:22
JPMaverick
311
New contributor
JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 6 at 18:22
JPMaverick
311
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JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 6 at 18:22
JPMaverick
311
asked Sep 6 at 18:22
JPMaverick
311
311
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JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
JPMaverick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
3 Answers
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up vote
8
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It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.
Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.
answered Sep 6 at 18:47
Nuclear Wang
2,239818
Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
– JPMaverick
Sep 6 at 18:59
The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
– Nuclear Wang
Sep 6 at 19:29
Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
– JPMaverick
Sep 6 at 19:36
I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
– JPMaverick
Sep 6 at 19:52
1
@JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
– Nuclear Wang
Sep 6 at 19:59
 |Â
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3
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Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
$$
fracacdot b2frac12a cdot 2b = frac18.
$$
answered Sep 6 at 22:33
katosh
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Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.
For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.
Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.
- The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)
- The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$
- etc.
Putting each of those terms into a summation, we get:
$(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$
Or,
$sum_i=0^n-1 fracin^2$
How to handle the fact A and B do not deliver the same number of packages?
I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.
How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.
$frac12 cdot sum_i=0^b-1 fracib^2$
answered yesterday
Tommy
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The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
– whuber♦
yesterday
You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
– Tommy
yesterday
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.
Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.
answered Sep 6 at 18:47
Nuclear Wang
2,239818
Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
– JPMaverick
Sep 6 at 18:59
The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
– Nuclear Wang
Sep 6 at 19:29
Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
– JPMaverick
Sep 6 at 19:36
I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
– JPMaverick
Sep 6 at 19:52
1
@JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
– Nuclear Wang
Sep 6 at 19:59
 |Â
show 3 more comments
up vote
8
down vote
It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.
Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.
answered Sep 6 at 18:47
Nuclear Wang
2,239818
Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
– JPMaverick
Sep 6 at 18:59
The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
– Nuclear Wang
Sep 6 at 19:29
Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
– JPMaverick
Sep 6 at 19:36
I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
– JPMaverick
Sep 6 at 19:52
1
@JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
– Nuclear Wang
Sep 6 at 19:59
 |Â
show 3 more comments
up vote
8
down vote
up vote
8
down vote
It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.
Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.
answered Sep 6 at 18:47
Nuclear Wang
2,239818
It's 1/8. See the figure below, which shows A's delivery time on the x-axis and B's on the y-axis. Since deliveries are uniformly distributed, all points in the square are equally likely to occur. B delivers before A only in the shaded region, which is 1/8 of the total figure.
Another way to think of it is that there's a 50% chance A delivers before B even starts, and 50% chance that B delivers after A is done, meaning there's a 75% chance of one or both of those happening. In the 25% chance they both deliver in the overlapping hour, it's a 50-50 chance of which delivers first.
answered Sep 6 at 18:47
Nuclear Wang
2,239818
edited Sep 6 at 18:52
answered Sep 6 at 18:47
Nuclear Wang
2,239818
answered Sep 6 at 18:47
Nuclear Wang
2,239818
answered Sep 6 at 18:47
Nuclear Wang
2,239818
2,239818
Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
– JPMaverick
Sep 6 at 18:59
The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
– Nuclear Wang
Sep 6 at 19:29
Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
– JPMaverick
Sep 6 at 19:36
I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
– JPMaverick
Sep 6 at 19:52
1
@JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
– Nuclear Wang
Sep 6 at 19:59
 |Â
show 3 more comments
Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
– JPMaverick
Sep 6 at 18:59
The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
– Nuclear Wang
Sep 6 at 19:29
Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
– JPMaverick
Sep 6 at 19:36
I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
– JPMaverick
Sep 6 at 19:52
1
@JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
– Nuclear Wang
Sep 6 at 19:59
Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
– JPMaverick
Sep 6 at 18:59
Thanks for the answer and especially the graph, and that is the answer I eventually gave (using your "another way to think about it"). But since the way the question was posed is referring only to the trips made by B, it seems to me that it's actually 1/4, since for every 100 trips made by B, 25 of them are before A in the overlapping hour. In other words, why does the portion where there was no chance that B could arrive first because B was not making any trips count? I read the question as only asking about B's trips. But my probability skills are admittedly weak after years of disuse.
– JPMaverick
Sep 6 at 18:59
The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
– Nuclear Wang
Sep 6 at 19:29
The answer holds because it's asking about "any delivery from B" compared to "any delivery from A". Imagine a case where A delivers from 6am to 10am instead of 8am-10am. Now it's even less likely that a random delivery from B is performed before a random delivery from A. If you don't count A's deliveries that occur outside of B's hours, extending A's delivery window to be earlier would have no effect.
– Nuclear Wang
Sep 6 at 19:29
Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
– JPMaverick
Sep 6 at 19:36
Perfect, got it. Thanks so much. Guess I won't be getting admitted into THAT program, lol
– JPMaverick
Sep 6 at 19:36
I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
– JPMaverick
Sep 6 at 19:52
I still can't seem to let this go, because it still seems to me that A's window getting extended is irrelevant if the question is "Of all of B's trips, what percentage of them took place before at least one of A's trips". Since the distribution is uniform, the number of potentially overlapping trips taken does not change by extending A's delivery window. It seems that there is actually NO answer possible without knowing the delivery rates of both A and B, if the questions is posed only about B (as it was). Sorry for being dense, if I am, and perfectly understandable if this is getting boring
– JPMaverick
Sep 6 at 19:52
1
1
@JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
– Nuclear Wang
Sep 6 at 19:59
@JPMaverick, I agree the exact wording of the question is important. This answer works when it's worded as "any given delivery from B" and "any given delivery from A". If it compares "any given delivery from B" to "any delivery from A", we need to know how many deliveries A makes. If A makes a million deliveries between 8 and 10, it's extremely unlikely that a random delivery from B occurs before A has delivered anything. If A only makes 2 deliveries in that time, it's reasonably possible that "any given delivery from B" occurs before "any" delivery from A (i.e. the first delivery from A).
– Nuclear Wang
Sep 6 at 19:59
 |Â
show 3 more comments
up vote
3
down vote
Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
$$
fracacdot b2frac12a cdot 2b = frac18.
$$
answered Sep 6 at 22:33
katosh
1093
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katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote
Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
$$
fracacdot b2frac12a cdot 2b = frac18.
$$
answered Sep 6 at 22:33
katosh
1093
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katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote
up vote
3
down vote
Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
$$
fracacdot b2frac12a cdot 2b = frac18.
$$
answered Sep 6 at 22:33
katosh
1093
New contributor
katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Since the delivery rates are not specified, lets assume A delivers $a$ packages per hour and B delivers $b$ packages per hour. So there are $2a cdot 2b$ pairs of delivery times. The window in which A and B overlap in deliver times has only $a cdot b$ pairs, in half of which A comes before B. So the proportion of pairs in which A comes before B is
$$
fracacdot b2frac12a cdot 2b = frac18.
$$
answered Sep 6 at 22:33
katosh
1093
New contributor
katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Sep 6 at 22:33
katosh
1093
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answered Sep 6 at 22:33
katosh
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answered Sep 6 at 22:33
katosh
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1093
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New contributor
katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
katosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.
For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.
Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.
- The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)
- The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$
- etc.
Putting each of those terms into a summation, we get:
$(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$
Or,
$sum_i=0^n-1 fracin^2$
How to handle the fact A and B do not deliver the same number of packages?
I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.
How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.
$frac12 cdot sum_i=0^b-1 fracib^2$
answered yesterday
Tommy
11
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The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
– whuber♦
yesterday
You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
– Tommy
yesterday
add a comment |Â
up vote
0
down vote
Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.
For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.
Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.
- The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)
- The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$
- etc.
Putting each of those terms into a summation, we get:
$(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$
Or,
$sum_i=0^n-1 fracin^2$
How to handle the fact A and B do not deliver the same number of packages?
I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.
How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.
$frac12 cdot sum_i=0^b-1 fracib^2$
answered yesterday
Tommy
11
New contributor
Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
– whuber♦
yesterday
You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
– Tommy
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.
For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.
Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.
- The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)
- The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$
- etc.
Putting each of those terms into a summation, we get:
$(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$
Or,
$sum_i=0^n-1 fracin^2$
How to handle the fact A and B do not deliver the same number of packages?
I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.
How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.
$frac12 cdot sum_i=0^b-1 fracib^2$
answered yesterday
Tommy
11
New contributor
Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Stumbled across this and it got in my head. :-)
The answer seems like it must depend on the relative number of deliveries each truck makes in the hour of possible overlap (9a-10a) -- there's no constant answer.
For example, suppose each truck makes 2 total deliveries (1 per hour). They'd each make 1 delivery between 9 and 10 and B wouldn't beat anything from A. So, the probability is 0 in that case.
Consider a simplified version of the problem where they both only make deliveries between 9-10a (still a uniform distribution). And, for starters, suppose they make the same number of deliveries, n.
- The first delivery for B will beat everything except the first delivery from A (which it ties). So, with probability $frac1n$ (the probability we're the first delivery for B) we beat an event with probability $fracn-1n$ (the probability we're not the first delivery of A)
- The second delivery for B will beat everything except the first two deliveries from A. So, with probability $frac1n$ we beat an event with probablity $fracn-2n$
- etc.
Putting each of those terms into a summation, we get:
$(frac1n cdot fracn-1n) + (frac1n cdot fracn-2n) + ... + (frac1n cdot fracn-nn)$
Or,
$sum_i=0^n-1 fracin^2$
How to handle the fact A and B do not deliver the same number of packages?
I'm not sure, to be honest. I think for large values of A and B, it will be similar to the above, because each delivery for B will "beat" a fraction of A similar to as if they were equal, but this breaks down for smaller numbers.
How to handle the fact that A and B deliver over a 2 hour period and not just the one hour period of interest? I think, because these are uniform over the entire period, we can scale the probability by $frac12$. So, I see the final answer being something like this, where $b$ is the total number of packages truck B delivers.
$frac12 cdot sum_i=0^b-1 fracib^2$
answered yesterday
Tommy
11
New contributor
Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
answered yesterday
Tommy
11
New contributor
Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Tommy
11
answered yesterday
Tommy
11
11
New contributor
Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tommy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
– whuber♦
yesterday
You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
– Tommy
yesterday
add a comment |Â
The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
– whuber♦
yesterday
You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
– Tommy
yesterday
The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
– whuber♦
yesterday
The question asks for a probability. It is difficult to find any realistic values of $a,b,$ and $n$ for which your formulas produce results between $0$ and $1,$ so they must be wrong. As other respondents have shown, there is a unique answer. It derives from the assumption of a uniform probability distribution for the delivery time for each truck.
– whuber♦
yesterday
You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
– Tommy
yesterday
You're absolutely right. I goofed horribly in a few places. I updated the answer and I think it sanity checks for a few small values.
– Tommy
yesterday
add a comment |Â
JPMaverick is a new contributor. Be nice, and check out our Code of Conduct.
JPMaverick is a new contributor. Be nice, and check out our Code of Conduct.
JPMaverick is a new contributor. Be nice, and check out our Code of Conduct.
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