Probability of objects are being sorted
$begingroup$
There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
$endgroup$
add a comment |
$begingroup$
There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
$endgroup$
add a comment |
$begingroup$
There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
$endgroup$
There are 20 toffees (10 of mint, 10 of orange) in two boxes. I mix them together. Then I put 10 randomly chosen from the pot into one box and 10 in the other box. What is the probability that I end up with 10 orange toffees in one box and 10 mint toffees in the other?
probability probability-theory elementary-probability
probability probability-theory elementary-probability
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
asked Dec 3 '18 at 17:58
Maths_RocksMaths_Rocks
30217
30217
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 '18 at 18:04
OfyaOfya
5198
$endgroup$
add a comment |
$begingroup$
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
$endgroup$
$begingroup$
Shouldn't answer be simply 1/2
$endgroup$
– Maths_Rocks
Dec 4 '18 at 1:09
$begingroup$
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
$endgroup$
– Sean Lee
Dec 4 '18 at 2:36
$begingroup$
Yes got my mistake...thanx !!
$endgroup$
– Maths_Rocks
Dec 4 '18 at 9:03
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%2f3024440%2fprobability-of-objects-are-being-sorted%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
$begingroup$
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 '18 at 18:04
OfyaOfya
5198
$endgroup$
add a comment |
$begingroup$
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 '18 at 18:04
OfyaOfya
5198
$endgroup$
add a comment |
$begingroup$
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 '18 at 18:04
OfyaOfya
5198
$endgroup$
There is only 2 possible ways to get the desired arrangement. Mint in box A, rest in box B or Orange in box A and mint in box B.
The total number of ways you can randomly put is $frac {20!} {10! 10!}$ or $20 choose 10$.
So the probability is $frac 2 {20 choose 10}$ or $frac {2. 10!. 10!} {20!}$
answered Dec 3 '18 at 18:04
OfyaOfya
5198
answered Dec 3 '18 at 18:04
OfyaOfya
5198
answered Dec 3 '18 at 18:04
OfyaOfya
5198
answered Dec 3 '18 at 18:04
OfyaOfya
5198
5198
add a comment |
add a comment |
$begingroup$
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
$endgroup$
$begingroup$
Shouldn't answer be simply 1/2
$endgroup$
– Maths_Rocks
Dec 4 '18 at 1:09
$begingroup$
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
$endgroup$
– Sean Lee
Dec 4 '18 at 2:36
$begingroup$
Yes got my mistake...thanx !!
$endgroup$
– Maths_Rocks
Dec 4 '18 at 9:03
add a comment |
$begingroup$
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
$endgroup$
$begingroup$
Shouldn't answer be simply 1/2
$endgroup$
– Maths_Rocks
Dec 4 '18 at 1:09
$begingroup$
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
$endgroup$
– Sean Lee
Dec 4 '18 at 2:36
$begingroup$
Yes got my mistake...thanx !!
$endgroup$
– Maths_Rocks
Dec 4 '18 at 9:03
add a comment |
$begingroup$
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
$endgroup$
You can view this question via the following process.
First, choose 10 toffees - there are $20 choose 10$ such choices and put them into one box. Put the remaining toffees in the other box.
There are only two scenarios where you get the desired outcome (perfectly separated), i.e. choosing all 10 orange first, or choosing all 10 mint first, thus you would have:
$$
Pr(text{Perfectly Separated}) = frac{2}{20 choose 10}
$$
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
answered Dec 3 '18 at 18:06
Sean LeeSean Lee
393111
393111
$begingroup$
Shouldn't answer be simply 1/2
$endgroup$
– Maths_Rocks
Dec 4 '18 at 1:09
$begingroup$
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
$endgroup$
– Sean Lee
Dec 4 '18 at 2:36
$begingroup$
Yes got my mistake...thanx !!
$endgroup$
– Maths_Rocks
Dec 4 '18 at 9:03
add a comment |
$begingroup$
Shouldn't answer be simply 1/2
$endgroup$
– Maths_Rocks
Dec 4 '18 at 1:09
$begingroup$
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
$endgroup$
– Sean Lee
Dec 4 '18 at 2:36
$begingroup$
Yes got my mistake...thanx !!
$endgroup$
– Maths_Rocks
Dec 4 '18 at 9:03
$begingroup$
Shouldn't answer be simply 1/2
$endgroup$
– Maths_Rocks
Dec 4 '18 at 1:09
$begingroup$
Shouldn't answer be simply 1/2
$endgroup$
– Maths_Rocks
Dec 4 '18 at 1:09
$begingroup$
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
$endgroup$
– Sean Lee
Dec 4 '18 at 2:36
$begingroup$
No - try to think of this concretely. If you randomly pick 10 toffees, what is the chance that all of them will be the same colour?
$endgroup$
– Sean Lee
Dec 4 '18 at 2:36
$begingroup$
Yes got my mistake...thanx !!
$endgroup$
– Maths_Rocks
Dec 4 '18 at 9:03
$begingroup$
Yes got my mistake...thanx !!
$endgroup$
– Maths_Rocks
Dec 4 '18 at 9:03
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%2f3024440%2fprobability-of-objects-are-being-sorted%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
