What is the probability that the girl who laughed loudly was from room number 2?
$begingroup$
There are 4 boys and 2 girls in room number 1 and 5 boys and 3 girls in room number 2. A girl from one of the two rooms laughed loudly. What is the probability that the girl who laughed was from room number 2.
I am finding this question very confusing. One thing that I have figured out is we have to use Bayes' theorem.
If I take $E_1, E_2$ and $A$ as following:
$E_1=$Event in which the girl is from room number 1,
$E_2=$Event in which the girl is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
Then, we have to find $P(E_2/A)$
$P(E_1)=1/7$
$P(E_2)=3/14$
$P(A/E_1)=1/3$
$P(A/E_2)=3/8$
$P(E_2/A)=frac{P(A/E_2)P(E_2)}{P(A/E_1)P(E_1)+P(A/E_2)P(E_2)}=27/43$
If I consider $E_1, E_2$ and $A$ as the following,
$E_1=$Event in the person is from room number 1,
$E_2=$Event in the person is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
then I am getting 3/5.
Which one is correct?
probability bayesian bayes-theorem
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
$endgroup$
add a comment |
$begingroup$
There are 4 boys and 2 girls in room number 1 and 5 boys and 3 girls in room number 2. A girl from one of the two rooms laughed loudly. What is the probability that the girl who laughed was from room number 2.
I am finding this question very confusing. One thing that I have figured out is we have to use Bayes' theorem.
If I take $E_1, E_2$ and $A$ as following:
$E_1=$Event in which the girl is from room number 1,
$E_2=$Event in which the girl is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
Then, we have to find $P(E_2/A)$
$P(E_1)=1/7$
$P(E_2)=3/14$
$P(A/E_1)=1/3$
$P(A/E_2)=3/8$
$P(E_2/A)=frac{P(A/E_2)P(E_2)}{P(A/E_1)P(E_1)+P(A/E_2)P(E_2)}=27/43$
If I consider $E_1, E_2$ and $A$ as the following,
$E_1=$Event in the person is from room number 1,
$E_2=$Event in the person is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
then I am getting 3/5.
Which one is correct?
probability bayesian bayes-theorem
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
$endgroup$
1
$begingroup$
Are you sure the question isn't trying to trick you into using Bayes' theorem by giving you irrelevant information (namely the number of boys, unless you've got some information about how laughter - provoking they are)?
$endgroup$
– timtfj
Dec 25 '18 at 19:10
$begingroup$
This question was given in the exercise for Bayes' theorem. Many such questions can be solved without using the theorem directly but Bayes' theorem provides an organized approach as per my knowledge. If you use Bayes' theorem then you have to take the number of boys into account to get 3/5 which @Daniel Mathias also got. Also, you do not need information about "how laughter-provoking they are". The language of the question is a bit tricky. They are actually asking you to find the probability that a person is from room number 2 given that she is a girl.
$endgroup$
– MrAP
Dec 25 '18 at 20:52
add a comment |
$begingroup$
There are 4 boys and 2 girls in room number 1 and 5 boys and 3 girls in room number 2. A girl from one of the two rooms laughed loudly. What is the probability that the girl who laughed was from room number 2.
I am finding this question very confusing. One thing that I have figured out is we have to use Bayes' theorem.
If I take $E_1, E_2$ and $A$ as following:
$E_1=$Event in which the girl is from room number 1,
$E_2=$Event in which the girl is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
Then, we have to find $P(E_2/A)$
$P(E_1)=1/7$
$P(E_2)=3/14$
$P(A/E_1)=1/3$
$P(A/E_2)=3/8$
$P(E_2/A)=frac{P(A/E_2)P(E_2)}{P(A/E_1)P(E_1)+P(A/E_2)P(E_2)}=27/43$
If I consider $E_1, E_2$ and $A$ as the following,
$E_1=$Event in the person is from room number 1,
$E_2=$Event in the person is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
then I am getting 3/5.
Which one is correct?
probability bayesian bayes-theorem
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
$endgroup$
There are 4 boys and 2 girls in room number 1 and 5 boys and 3 girls in room number 2. A girl from one of the two rooms laughed loudly. What is the probability that the girl who laughed was from room number 2.
I am finding this question very confusing. One thing that I have figured out is we have to use Bayes' theorem.
If I take $E_1, E_2$ and $A$ as following:
$E_1=$Event in which the girl is from room number 1,
$E_2=$Event in which the girl is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
Then, we have to find $P(E_2/A)$
$P(E_1)=1/7$
$P(E_2)=3/14$
$P(A/E_1)=1/3$
$P(A/E_2)=3/8$
$P(E_2/A)=frac{P(A/E_2)P(E_2)}{P(A/E_1)P(E_1)+P(A/E_2)P(E_2)}=27/43$
If I consider $E_1, E_2$ and $A$ as the following,
$E_1=$Event in the person is from room number 1,
$E_2=$Event in the person is from room number 2,
$A=$Event in which a girl from one of the two rooms laughed loudly.
then I am getting 3/5.
Which one is correct?
probability bayesian bayes-theorem
probability bayesian bayes-theorem
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
edited Dec 25 '18 at 20:55
MrAP
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
asked Dec 25 '18 at 17:35
MrAPMrAP
1,15021432
1,15021432
1
$begingroup$
Are you sure the question isn't trying to trick you into using Bayes' theorem by giving you irrelevant information (namely the number of boys, unless you've got some information about how laughter - provoking they are)?
$endgroup$
– timtfj
Dec 25 '18 at 19:10
$begingroup$
This question was given in the exercise for Bayes' theorem. Many such questions can be solved without using the theorem directly but Bayes' theorem provides an organized approach as per my knowledge. If you use Bayes' theorem then you have to take the number of boys into account to get 3/5 which @Daniel Mathias also got. Also, you do not need information about "how laughter-provoking they are". The language of the question is a bit tricky. They are actually asking you to find the probability that a person is from room number 2 given that she is a girl.
$endgroup$
– MrAP
Dec 25 '18 at 20:52
add a comment |
1
$begingroup$
Are you sure the question isn't trying to trick you into using Bayes' theorem by giving you irrelevant information (namely the number of boys, unless you've got some information about how laughter - provoking they are)?
$endgroup$
– timtfj
Dec 25 '18 at 19:10
$begingroup$
This question was given in the exercise for Bayes' theorem. Many such questions can be solved without using the theorem directly but Bayes' theorem provides an organized approach as per my knowledge. If you use Bayes' theorem then you have to take the number of boys into account to get 3/5 which @Daniel Mathias also got. Also, you do not need information about "how laughter-provoking they are". The language of the question is a bit tricky. They are actually asking you to find the probability that a person is from room number 2 given that she is a girl.
$endgroup$
– MrAP
Dec 25 '18 at 20:52
1
1
$begingroup$
Are you sure the question isn't trying to trick you into using Bayes' theorem by giving you irrelevant information (namely the number of boys, unless you've got some information about how laughter - provoking they are)?
$endgroup$
– timtfj
Dec 25 '18 at 19:10
$begingroup$
Are you sure the question isn't trying to trick you into using Bayes' theorem by giving you irrelevant information (namely the number of boys, unless you've got some information about how laughter - provoking they are)?
$endgroup$
– timtfj
Dec 25 '18 at 19:10
$begingroup$
This question was given in the exercise for Bayes' theorem. Many such questions can be solved without using the theorem directly but Bayes' theorem provides an organized approach as per my knowledge. If you use Bayes' theorem then you have to take the number of boys into account to get 3/5 which @Daniel Mathias also got. Also, you do not need information about "how laughter-provoking they are". The language of the question is a bit tricky. They are actually asking you to find the probability that a person is from room number 2 given that she is a girl.
$endgroup$
– MrAP
Dec 25 '18 at 20:52
$begingroup$
This question was given in the exercise for Bayes' theorem. Many such questions can be solved without using the theorem directly but Bayes' theorem provides an organized approach as per my knowledge. If you use Bayes' theorem then you have to take the number of boys into account to get 3/5 which @Daniel Mathias also got. Also, you do not need information about "how laughter-provoking they are". The language of the question is a bit tricky. They are actually asking you to find the probability that a person is from room number 2 given that she is a girl.
$endgroup$
– MrAP
Dec 25 '18 at 20:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A girl laughed.
There are $5$ girls.
$3$ of the $5$ are in room $2$.
$P=frac{3}{5}$
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
$endgroup$
$begingroup$
While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability).
$endgroup$
– Sambo
Dec 25 '18 at 18:05
2
$begingroup$
@sambo I disagree. I think it is perfectly rigorous.
$endgroup$
– Zachary Selk
Dec 25 '18 at 18:07
$begingroup$
@Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard.
$endgroup$
– Sambo
Dec 25 '18 at 18:18
1
$begingroup$
@Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant.
$endgroup$
– timtfj
Dec 25 '18 at 19:16
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%2f3052300%2fwhat-is-the-probability-that-the-girl-who-laughed-loudly-was-from-room-number-2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A girl laughed.
There are $5$ girls.
$3$ of the $5$ are in room $2$.
$P=frac{3}{5}$
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
$endgroup$
$begingroup$
While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability).
$endgroup$
– Sambo
Dec 25 '18 at 18:05
2
$begingroup$
@sambo I disagree. I think it is perfectly rigorous.
$endgroup$
– Zachary Selk
Dec 25 '18 at 18:07
$begingroup$
@Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard.
$endgroup$
– Sambo
Dec 25 '18 at 18:18
1
$begingroup$
@Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant.
$endgroup$
– timtfj
Dec 25 '18 at 19:16
add a comment |
$begingroup$
A girl laughed.
There are $5$ girls.
$3$ of the $5$ are in room $2$.
$P=frac{3}{5}$
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
$endgroup$
$begingroup$
While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability).
$endgroup$
– Sambo
Dec 25 '18 at 18:05
2
$begingroup$
@sambo I disagree. I think it is perfectly rigorous.
$endgroup$
– Zachary Selk
Dec 25 '18 at 18:07
$begingroup$
@Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard.
$endgroup$
– Sambo
Dec 25 '18 at 18:18
1
$begingroup$
@Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant.
$endgroup$
– timtfj
Dec 25 '18 at 19:16
add a comment |
$begingroup$
A girl laughed.
There are $5$ girls.
$3$ of the $5$ are in room $2$.
$P=frac{3}{5}$
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
$endgroup$
A girl laughed.
There are $5$ girls.
$3$ of the $5$ are in room $2$.
$P=frac{3}{5}$
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
answered Dec 25 '18 at 17:58
Daniel MathiasDaniel Mathias
1,29018
1,29018
$begingroup$
While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability).
$endgroup$
– Sambo
Dec 25 '18 at 18:05
2
$begingroup$
@sambo I disagree. I think it is perfectly rigorous.
$endgroup$
– Zachary Selk
Dec 25 '18 at 18:07
$begingroup$
@Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard.
$endgroup$
– Sambo
Dec 25 '18 at 18:18
1
$begingroup$
@Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant.
$endgroup$
– timtfj
Dec 25 '18 at 19:16
add a comment |
$begingroup$
While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability).
$endgroup$
– Sambo
Dec 25 '18 at 18:05
2
$begingroup$
@sambo I disagree. I think it is perfectly rigorous.
$endgroup$
– Zachary Selk
Dec 25 '18 at 18:07
$begingroup$
@Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard.
$endgroup$
– Sambo
Dec 25 '18 at 18:18
1
$begingroup$
@Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant.
$endgroup$
– timtfj
Dec 25 '18 at 19:16
$begingroup$
While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability).
$endgroup$
– Sambo
Dec 25 '18 at 18:05
$begingroup$
While I agree with your answer, you do not provide a very rigorous reasoning (in particular, no mention of events and conditional probability).
$endgroup$
– Sambo
Dec 25 '18 at 18:05
2
2
$begingroup$
@sambo I disagree. I think it is perfectly rigorous.
$endgroup$
– Zachary Selk
Dec 25 '18 at 18:07
$begingroup$
@sambo I disagree. I think it is perfectly rigorous.
$endgroup$
– Zachary Selk
Dec 25 '18 at 18:07
$begingroup$
@Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard.
$endgroup$
– Sambo
Dec 25 '18 at 18:18
$begingroup$
@Zachary Perhaps rigorous was not the right word. I meant the answer isn't formulated with usual probability theory vocabulary, like the question was. It seems like OP wants to produce an answer this way, and this explanation, while clear, doesn't help in that regard.
$endgroup$
– Sambo
Dec 25 '18 at 18:18
1
1
$begingroup$
@Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant.
$endgroup$
– timtfj
Dec 25 '18 at 19:16
$begingroup$
@Sambo I think the problem is meant as an exercise in ignoring irrelevant information (the number of boys)—or is maybe followed by other versions in which the boys become relevant.
$endgroup$
– timtfj
Dec 25 '18 at 19:16
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%2f3052300%2fwhat-is-the-probability-that-the-girl-who-laughed-loudly-was-from-room-number-2%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
$begingroup$
Are you sure the question isn't trying to trick you into using Bayes' theorem by giving you irrelevant information (namely the number of boys, unless you've got some information about how laughter - provoking they are)?
$endgroup$
– timtfj
Dec 25 '18 at 19:10
$begingroup$
This question was given in the exercise for Bayes' theorem. Many such questions can be solved without using the theorem directly but Bayes' theorem provides an organized approach as per my knowledge. If you use Bayes' theorem then you have to take the number of boys into account to get 3/5 which @Daniel Mathias also got. Also, you do not need information about "how laughter-provoking they are". The language of the question is a bit tricky. They are actually asking you to find the probability that a person is from room number 2 given that she is a girl.
$endgroup$
– MrAP
Dec 25 '18 at 20:52