Mixing
Average of shortest half of two halves and probability of three halves making a triangle
$begingroup$
This is a problem that has been bothering me because it seems so easy however the answers don't feel right so...(It showed up in my latest statistics exam and almost everybody got it wrong because we thought it was just too plain easy)
"A point is randomly chosen in the [0,1] interval splitting it in two halves."
1.Whats the average length of the shortest half?
Isn't the average always 0.5?
2.Now picking two random points in the same interval what's the probability of being able to create a triangle with the 3 segments?
Shouldn't this be 1?
probability geometric-probability
asked Feb 1 at 20:35
LeoLeo
82
$endgroup$
add a comment |
$begingroup$
This is a problem that has been bothering me because it seems so easy however the answers don't feel right so...(It showed up in my latest statistics exam and almost everybody got it wrong because we thought it was just too plain easy)
"A point is randomly chosen in the [0,1] interval splitting it in two halves."
1.Whats the average length of the shortest half?
Isn't the average always 0.5?
2.Now picking two random points in the same interval what's the probability of being able to create a triangle with the 3 segments?
Shouldn't this be 1?
probability geometric-probability
asked Feb 1 at 20:35
LeoLeo
82
$endgroup$
1
$begingroup$
They mean "spitting it into two segments". As you say, the segments probably won't have the same length.
$endgroup$
– lulu
Feb 1 at 20:36
$begingroup$
The three segments have to satisfy the triangle inequality: the longest must be less than the sum of the two others.
$endgroup$
– saulspatz
Feb 1 at 20:39
$begingroup$
The shortest piece in part a has a random length with uniform distribution over [0,1/2]. You need the expected value of that.
$endgroup$
– GReyes
Feb 1 at 20:49
add a comment |
$begingroup$
This is a problem that has been bothering me because it seems so easy however the answers don't feel right so...(It showed up in my latest statistics exam and almost everybody got it wrong because we thought it was just too plain easy)
"A point is randomly chosen in the [0,1] interval splitting it in two halves."
1.Whats the average length of the shortest half?
Isn't the average always 0.5?
2.Now picking two random points in the same interval what's the probability of being able to create a triangle with the 3 segments?
Shouldn't this be 1?
probability geometric-probability
asked Feb 1 at 20:35
LeoLeo
82
$endgroup$
This is a problem that has been bothering me because it seems so easy however the answers don't feel right so...(It showed up in my latest statistics exam and almost everybody got it wrong because we thought it was just too plain easy)
"A point is randomly chosen in the [0,1] interval splitting it in two halves."
1.Whats the average length of the shortest half?
Isn't the average always 0.5?
2.Now picking two random points in the same interval what's the probability of being able to create a triangle with the 3 segments?
Shouldn't this be 1?
probability geometric-probability
probability geometric-probability
asked Feb 1 at 20:35
LeoLeo
82
asked Feb 1 at 20:35
LeoLeo
82
edited Feb 1 at 21:11
Leo
asked Feb 1 at 20:35
LeoLeo
82
asked Feb 1 at 20:35
LeoLeo
82
asked Feb 1 at 20:35
LeoLeo
82
82
1
$begingroup$
They mean "spitting it into two segments". As you say, the segments probably won't have the same length.
$endgroup$
– lulu
Feb 1 at 20:36
$begingroup$
The three segments have to satisfy the triangle inequality: the longest must be less than the sum of the two others.
$endgroup$
– saulspatz
Feb 1 at 20:39
$begingroup$
The shortest piece in part a has a random length with uniform distribution over [0,1/2]. You need the expected value of that.
$endgroup$
– GReyes
Feb 1 at 20:49
add a comment |
1
$begingroup$
They mean "spitting it into two segments". As you say, the segments probably won't have the same length.
$endgroup$
– lulu
Feb 1 at 20:36
$begingroup$
The three segments have to satisfy the triangle inequality: the longest must be less than the sum of the two others.
$endgroup$
– saulspatz
Feb 1 at 20:39
$begingroup$
The shortest piece in part a has a random length with uniform distribution over [0,1/2]. You need the expected value of that.
$endgroup$
– GReyes
Feb 1 at 20:49
1
1
$begingroup$
They mean "spitting it into two segments". As you say, the segments probably won't have the same length.
$endgroup$
– lulu
Feb 1 at 20:36
$begingroup$
They mean "spitting it into two segments". As you say, the segments probably won't have the same length.
$endgroup$
– lulu
Feb 1 at 20:36
$begingroup$
The three segments have to satisfy the triangle inequality: the longest must be less than the sum of the two others.
$endgroup$
– saulspatz
Feb 1 at 20:39
$begingroup$
The three segments have to satisfy the triangle inequality: the longest must be less than the sum of the two others.
$endgroup$
– saulspatz
Feb 1 at 20:39
$begingroup$
The shortest piece in part a has a random length with uniform distribution over [0,1/2]. You need the expected value of that.
$endgroup$
– GReyes
Feb 1 at 20:49
$begingroup$
The shortest piece in part a has a random length with uniform distribution over [0,1/2]. You need the expected value of that.
$endgroup$
– GReyes
Feb 1 at 20:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $l$ be the length of the shortest half. Then
$$
mathsf{E}l=int_0^{1/2}x,dx+int_{1/2}^1 (1-x), dx = frac{1}{8}+frac{1}{8} = frac{1}{4}.
$$Look at this question (your probability is $1-$ the probability that the length of one of the segments is greater than $1/2$).
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
$endgroup$
add a comment |
$begingroup$
a) The short piece must be less than or equal to $0.5.$ So, it is not possible for the expected value to be $0.5$
$X$ is a unifrom random variable in $[0,1]$
$Y = min (x, 1-x)$
Find $E[Y]$
b)
In order for the 3 segments to form a triangle, they must stratify the "triangle inequality." That is the longest of the 3 segments must be shorter, than the sum of the other two segments. (i.e. it must be less than $0.5$)
This vid has a nice explanation and solution.
https://www.youtube.com/watch?v=udxwP26gTwA
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
$endgroup$
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',
autoActivateHeartbeat: false,
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%2f3096701%2faverage-of-shortest-half-of-two-halves-and-probability-of-three-halves-making-a%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
$begingroup$
Let $l$ be the length of the shortest half. Then
$$
mathsf{E}l=int_0^{1/2}x,dx+int_{1/2}^1 (1-x), dx = frac{1}{8}+frac{1}{8} = frac{1}{4}.
$$Look at this question (your probability is $1-$ the probability that the length of one of the segments is greater than $1/2$).
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
$endgroup$
add a comment |
$begingroup$
Let $l$ be the length of the shortest half. Then
$$
mathsf{E}l=int_0^{1/2}x,dx+int_{1/2}^1 (1-x), dx = frac{1}{8}+frac{1}{8} = frac{1}{4}.
$$Look at this question (your probability is $1-$ the probability that the length of one of the segments is greater than $1/2$).
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
$endgroup$
add a comment |
$begingroup$
Let $l$ be the length of the shortest half. Then
$$
mathsf{E}l=int_0^{1/2}x,dx+int_{1/2}^1 (1-x), dx = frac{1}{8}+frac{1}{8} = frac{1}{4}.
$$Look at this question (your probability is $1-$ the probability that the length of one of the segments is greater than $1/2$).
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
$endgroup$
Let $l$ be the length of the shortest half. Then
$$
mathsf{E}l=int_0^{1/2}x,dx+int_{1/2}^1 (1-x), dx = frac{1}{8}+frac{1}{8} = frac{1}{4}.
$$Look at this question (your probability is $1-$ the probability that the length of one of the segments is greater than $1/2$).
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
edited Feb 1 at 20:53
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
answered Feb 1 at 20:45
d.k.o.d.k.o.
10.6k730
10.6k730
add a comment |
add a comment |
$begingroup$
a) The short piece must be less than or equal to $0.5.$ So, it is not possible for the expected value to be $0.5$
$X$ is a unifrom random variable in $[0,1]$
$Y = min (x, 1-x)$
Find $E[Y]$
b)
In order for the 3 segments to form a triangle, they must stratify the "triangle inequality." That is the longest of the 3 segments must be shorter, than the sum of the other two segments. (i.e. it must be less than $0.5$)
This vid has a nice explanation and solution.
https://www.youtube.com/watch?v=udxwP26gTwA
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
a) The short piece must be less than or equal to $0.5.$ So, it is not possible for the expected value to be $0.5$
$X$ is a unifrom random variable in $[0,1]$
$Y = min (x, 1-x)$
Find $E[Y]$
b)
In order for the 3 segments to form a triangle, they must stratify the "triangle inequality." That is the longest of the 3 segments must be shorter, than the sum of the other two segments. (i.e. it must be less than $0.5$)
This vid has a nice explanation and solution.
https://www.youtube.com/watch?v=udxwP26gTwA
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
a) The short piece must be less than or equal to $0.5.$ So, it is not possible for the expected value to be $0.5$
$X$ is a unifrom random variable in $[0,1]$
$Y = min (x, 1-x)$
Find $E[Y]$
b)
In order for the 3 segments to form a triangle, they must stratify the "triangle inequality." That is the longest of the 3 segments must be shorter, than the sum of the other two segments. (i.e. it must be less than $0.5$)
This vid has a nice explanation and solution.
https://www.youtube.com/watch?v=udxwP26gTwA
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
$endgroup$
a) The short piece must be less than or equal to $0.5.$ So, it is not possible for the expected value to be $0.5$
$X$ is a unifrom random variable in $[0,1]$
$Y = min (x, 1-x)$
Find $E[Y]$
b)
In order for the 3 segments to form a triangle, they must stratify the "triangle inequality." That is the longest of the 3 segments must be shorter, than the sum of the other two segments. (i.e. it must be less than $0.5$)
This vid has a nice explanation and solution.
https://www.youtube.com/watch?v=udxwP26gTwA
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
edited Feb 1 at 20:55
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
answered Feb 1 at 20:45
Doug MDoug M
45.4k31954
45.4k31954
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.
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%2f3096701%2faverage-of-shortest-half-of-two-halves-and-probability-of-three-halves-making-a%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
1
$begingroup$
They mean "spitting it into two segments". As you say, the segments probably won't have the same length.
$endgroup$
– lulu
Feb 1 at 20:36
$begingroup$
The three segments have to satisfy the triangle inequality: the longest must be less than the sum of the two others.
$endgroup$
– saulspatz
Feb 1 at 20:39
$begingroup$
The shortest piece in part a has a random length with uniform distribution over [0,1/2]. You need the expected value of that.
$endgroup$
– GReyes
Feb 1 at 20:49