Probability for Uniform distributions
up vote
1
down vote
favorite
Two people decide to meet at a cafeteria sometime between 12:00 and 13:00. If, each and one of them arrive at random chosen times during the hour, and wait 45 minutes on each other (or until the clock strikes 13:00). What is the probability that they will meet?
My teacher told me that this can be solved drawing a quadrate, with I think 0 and 1 at each corner, respectively. Drawing a graph, and finding a cross-section would then yield the answer. The problem is I don't know how to do the graphs. He alse told me that the time each person waits may be regarded as a uniform distribution. So, my guess is that he means that
$$
t_1 in U(0,60), t_2 in U(0,60)
$$
But from here i don't know where to go. Anyone?
probability-theory uniform-distribution
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
add a comment |
up vote
1
down vote
favorite
Two people decide to meet at a cafeteria sometime between 12:00 and 13:00. If, each and one of them arrive at random chosen times during the hour, and wait 45 minutes on each other (or until the clock strikes 13:00). What is the probability that they will meet?
My teacher told me that this can be solved drawing a quadrate, with I think 0 and 1 at each corner, respectively. Drawing a graph, and finding a cross-section would then yield the answer. The problem is I don't know how to do the graphs. He alse told me that the time each person waits may be regarded as a uniform distribution. So, my guess is that he means that
$$
t_1 in U(0,60), t_2 in U(0,60)
$$
But from here i don't know where to go. Anyone?
probability-theory uniform-distribution
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
Similar to math.stackexchange.com/questions/103015/… or other questions linked to it
– Henry
Nov 26 at 18:51
The arrival time of each person is a uniform random variable like you mentioned. They will meet if the difference in arrival times is less than 45 minutes. So you need to calculate $P(|t_1-t_2|<45)$
– gd1035
Nov 26 at 18:51
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Two people decide to meet at a cafeteria sometime between 12:00 and 13:00. If, each and one of them arrive at random chosen times during the hour, and wait 45 minutes on each other (or until the clock strikes 13:00). What is the probability that they will meet?
My teacher told me that this can be solved drawing a quadrate, with I think 0 and 1 at each corner, respectively. Drawing a graph, and finding a cross-section would then yield the answer. The problem is I don't know how to do the graphs. He alse told me that the time each person waits may be regarded as a uniform distribution. So, my guess is that he means that
$$
t_1 in U(0,60), t_2 in U(0,60)
$$
But from here i don't know where to go. Anyone?
probability-theory uniform-distribution
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
Two people decide to meet at a cafeteria sometime between 12:00 and 13:00. If, each and one of them arrive at random chosen times during the hour, and wait 45 minutes on each other (or until the clock strikes 13:00). What is the probability that they will meet?
My teacher told me that this can be solved drawing a quadrate, with I think 0 and 1 at each corner, respectively. Drawing a graph, and finding a cross-section would then yield the answer. The problem is I don't know how to do the graphs. He alse told me that the time each person waits may be regarded as a uniform distribution. So, my guess is that he means that
$$
t_1 in U(0,60), t_2 in U(0,60)
$$
But from here i don't know where to go. Anyone?
probability-theory uniform-distribution
probability-theory uniform-distribution
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
asked Nov 26 at 18:40
Kristoffer Jerzy Linder
316
316
Similar to math.stackexchange.com/questions/103015/… or other questions linked to it
– Henry
Nov 26 at 18:51
The arrival time of each person is a uniform random variable like you mentioned. They will meet if the difference in arrival times is less than 45 minutes. So you need to calculate $P(|t_1-t_2|<45)$
– gd1035
Nov 26 at 18:51
add a comment |
Similar to math.stackexchange.com/questions/103015/… or other questions linked to it
– Henry
Nov 26 at 18:51
The arrival time of each person is a uniform random variable like you mentioned. They will meet if the difference in arrival times is less than 45 minutes. So you need to calculate $P(|t_1-t_2|<45)$
– gd1035
Nov 26 at 18:51
Similar to math.stackexchange.com/questions/103015/… or other questions linked to it
– Henry
Nov 26 at 18:51
Similar to math.stackexchange.com/questions/103015/… or other questions linked to it
– Henry
Nov 26 at 18:51
The arrival time of each person is a uniform random variable like you mentioned. They will meet if the difference in arrival times is less than 45 minutes. So you need to calculate $P(|t_1-t_2|<45)$
– gd1035
Nov 26 at 18:51
The arrival time of each person is a uniform random variable like you mentioned. They will meet if the difference in arrival times is less than 45 minutes. So you need to calculate $P(|t_1-t_2|<45)$
– gd1035
Nov 26 at 18:51
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Here is a plot of the starting time in minutes for $x$ (horizontal) and $y$ (vertical) and the cases where they will meet. Formally: $(x,y) s.t. |x-y| leq 45 wedge 0 leq x leq 60 wedge 0 leq y leq 60$.
The area of this region divided by the total possible region is the probability they will meet: $15/16$.
answered Nov 26 at 19:30
David G. Stork
9,33721232
Finally, I understand how to use the box. Thank You :-D
– Kristoffer Jerzy Linder
Nov 26 at 19:35
add a comment |
up vote
0
down vote
For simplicity, imagine they wait the full time regardless of whether they meet.
Anyone arriving after 12:15 will wait less than 45 minutes.
There's a $frac14$ chance for each of them that they'll actually wait 45 minutes.
Consider the cases where neither of them wait the full 45 minutes.
Trick question: What's the probability they meet?
Consider the cases where both of them wait the full 45 minutes.
Trick question: What's the probability they meet?
In the remaining cases, one of their arrivals is uniformly distributed from 12:00 to 12:15; consider that as a distribution of time earlier than 12:15 that they arrive.
The other arrival is uniformly distributed from 12:15 to 13:00; consider that as a distribution of time after 12:15 that they arrive.
They'll meet if the sum of those two times is less than 45 minutes.
I don't know what a quadrate is in this context, but it sounds like you have the tools you need to figure out the probability of that sum being less than 45 minutes, and to combine all the cases into a net probability.
answered Nov 26 at 19:04
ShapeOfMatter
313
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014741%2fprobability-for-uniform-distributions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here is a plot of the starting time in minutes for $x$ (horizontal) and $y$ (vertical) and the cases where they will meet. Formally: $(x,y) s.t. |x-y| leq 45 wedge 0 leq x leq 60 wedge 0 leq y leq 60$.
The area of this region divided by the total possible region is the probability they will meet: $15/16$.
answered Nov 26 at 19:30
David G. Stork
9,33721232
Finally, I understand how to use the box. Thank You :-D
– Kristoffer Jerzy Linder
Nov 26 at 19:35
add a comment |
up vote
1
down vote
Here is a plot of the starting time in minutes for $x$ (horizontal) and $y$ (vertical) and the cases where they will meet. Formally: $(x,y) s.t. |x-y| leq 45 wedge 0 leq x leq 60 wedge 0 leq y leq 60$.
The area of this region divided by the total possible region is the probability they will meet: $15/16$.
answered Nov 26 at 19:30
David G. Stork
9,33721232
Finally, I understand how to use the box. Thank You :-D
– Kristoffer Jerzy Linder
Nov 26 at 19:35
add a comment |
up vote
1
down vote
up vote
1
down vote
Here is a plot of the starting time in minutes for $x$ (horizontal) and $y$ (vertical) and the cases where they will meet. Formally: $(x,y) s.t. |x-y| leq 45 wedge 0 leq x leq 60 wedge 0 leq y leq 60$.
The area of this region divided by the total possible region is the probability they will meet: $15/16$.
answered Nov 26 at 19:30
David G. Stork
9,33721232
Here is a plot of the starting time in minutes for $x$ (horizontal) and $y$ (vertical) and the cases where they will meet. Formally: $(x,y) s.t. |x-y| leq 45 wedge 0 leq x leq 60 wedge 0 leq y leq 60$.
The area of this region divided by the total possible region is the probability they will meet: $15/16$.
answered Nov 26 at 19:30
David G. Stork
9,33721232
edited Nov 26 at 19:47
answered Nov 26 at 19:30
David G. Stork
9,33721232
answered Nov 26 at 19:30
David G. Stork
9,33721232
answered Nov 26 at 19:30
David G. Stork
9,33721232
9,33721232
Finally, I understand how to use the box. Thank You :-D
– Kristoffer Jerzy Linder
Nov 26 at 19:35
add a comment |
Finally, I understand how to use the box. Thank You :-D
– Kristoffer Jerzy Linder
Nov 26 at 19:35
Finally, I understand how to use the box. Thank You :-D
– Kristoffer Jerzy Linder
Nov 26 at 19:35
Finally, I understand how to use the box. Thank You :-D
– Kristoffer Jerzy Linder
Nov 26 at 19:35
add a comment |
up vote
0
down vote
For simplicity, imagine they wait the full time regardless of whether they meet.
Anyone arriving after 12:15 will wait less than 45 minutes.
There's a $frac14$ chance for each of them that they'll actually wait 45 minutes.
Consider the cases where neither of them wait the full 45 minutes.
Trick question: What's the probability they meet?
Consider the cases where both of them wait the full 45 minutes.
Trick question: What's the probability they meet?
In the remaining cases, one of their arrivals is uniformly distributed from 12:00 to 12:15; consider that as a distribution of time earlier than 12:15 that they arrive.
The other arrival is uniformly distributed from 12:15 to 13:00; consider that as a distribution of time after 12:15 that they arrive.
They'll meet if the sum of those two times is less than 45 minutes.
I don't know what a quadrate is in this context, but it sounds like you have the tools you need to figure out the probability of that sum being less than 45 minutes, and to combine all the cases into a net probability.
answered Nov 26 at 19:04
ShapeOfMatter
313
add a comment |
up vote
0
down vote
For simplicity, imagine they wait the full time regardless of whether they meet.
Anyone arriving after 12:15 will wait less than 45 minutes.
There's a $frac14$ chance for each of them that they'll actually wait 45 minutes.
Consider the cases where neither of them wait the full 45 minutes.
Trick question: What's the probability they meet?
Consider the cases where both of them wait the full 45 minutes.
Trick question: What's the probability they meet?
In the remaining cases, one of their arrivals is uniformly distributed from 12:00 to 12:15; consider that as a distribution of time earlier than 12:15 that they arrive.
The other arrival is uniformly distributed from 12:15 to 13:00; consider that as a distribution of time after 12:15 that they arrive.
They'll meet if the sum of those two times is less than 45 minutes.
I don't know what a quadrate is in this context, but it sounds like you have the tools you need to figure out the probability of that sum being less than 45 minutes, and to combine all the cases into a net probability.
answered Nov 26 at 19:04
ShapeOfMatter
313
add a comment |
up vote
0
down vote
up vote
0
down vote
For simplicity, imagine they wait the full time regardless of whether they meet.
Anyone arriving after 12:15 will wait less than 45 minutes.
There's a $frac14$ chance for each of them that they'll actually wait 45 minutes.
Consider the cases where neither of them wait the full 45 minutes.
Trick question: What's the probability they meet?
Consider the cases where both of them wait the full 45 minutes.
Trick question: What's the probability they meet?
In the remaining cases, one of their arrivals is uniformly distributed from 12:00 to 12:15; consider that as a distribution of time earlier than 12:15 that they arrive.
The other arrival is uniformly distributed from 12:15 to 13:00; consider that as a distribution of time after 12:15 that they arrive.
They'll meet if the sum of those two times is less than 45 minutes.
I don't know what a quadrate is in this context, but it sounds like you have the tools you need to figure out the probability of that sum being less than 45 minutes, and to combine all the cases into a net probability.
answered Nov 26 at 19:04
ShapeOfMatter
313
For simplicity, imagine they wait the full time regardless of whether they meet.
Anyone arriving after 12:15 will wait less than 45 minutes.
There's a $frac14$ chance for each of them that they'll actually wait 45 minutes.
Consider the cases where neither of them wait the full 45 minutes.
Trick question: What's the probability they meet?
Consider the cases where both of them wait the full 45 minutes.
Trick question: What's the probability they meet?
In the remaining cases, one of their arrivals is uniformly distributed from 12:00 to 12:15; consider that as a distribution of time earlier than 12:15 that they arrive.
The other arrival is uniformly distributed from 12:15 to 13:00; consider that as a distribution of time after 12:15 that they arrive.
They'll meet if the sum of those two times is less than 45 minutes.
I don't know what a quadrate is in this context, but it sounds like you have the tools you need to figure out the probability of that sum being less than 45 minutes, and to combine all the cases into a net probability.
answered Nov 26 at 19:04
ShapeOfMatter
313
answered Nov 26 at 19:04
ShapeOfMatter
313
answered Nov 26 at 19:04
ShapeOfMatter
313
answered Nov 26 at 19:04
ShapeOfMatter
313
313
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014741%2fprobability-for-uniform-distributions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Similar to math.stackexchange.com/questions/103015/… or other questions linked to it
– Henry
Nov 26 at 18:51
The arrival time of each person is a uniform random variable like you mentioned. They will meet if the difference in arrival times is less than 45 minutes. So you need to calculate $P(|t_1-t_2|<45)$
– gd1035
Nov 26 at 18:51