Usbrth

Inflated grade on resume at previous job, might former employer tell new employer?

What is the meaning of Triage in Cybersec world?

The difference between dialogue marks

Protecting Dualbooting Windows from dangerous code (like rm -rf)

Why don't Unix/Linux systems traverse through directories until they find the required version of a linked library?

I looked up a future colleague on LinkedIn before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?

I see my dog run

How long do I have to send my income tax payment to the IRS?

Why can Shazam do this?

How to deal with fear of taking dependencies

Why is my p-value correlated to difference between means in two sample tests?

What do hard-Brexiteers want with respect to the Irish border?

What is this 4-propeller plane?

Understanding the implication of what "well-defined" means for the operation in quotient group

Should I write numbers in words or as numerals when there are multiple next to each other?

What does "sndry explns" mean in one of the Hitchhiker's guide books?

Is there a name of the flying bionic bird?

How do you say "canon" as in "official for a story universe"?

How can I fix this gap between bookcases I made?

What can other administrators access on my machine?

How to change the limits of integration

What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?

Limit to 0 ambiguity

A poker game description that does not feel gimmicky

Does this elementary probability theory problem make sense?

0

$begingroup$

The problem forumaltes as:

Student takes test of 12 questions, each question is single-choice with 4 options. For each question he answers correctly if he knows the question, otherwise he guesses uniformly. Assume student answered correctly for exactly 9 questions. What is probability that he knew at least 7 questions?
My question is the following: does this problem statement make sense unless we have a priori probability distribution of how many questions does student know?

If it makes sense, how to solve it, and if it doesn't, how to show that and what could be the ways to complete the statement of this problem so that it would make sense?

probability-theory bayes-theorem

share|cite|improve this question

edited Apr 6 at 16:54

Bernard

124k741117

asked Apr 6 at 16:47

Mikhail KuzinMikhail Kuzin

304

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This question does make sense, assuming the Bayesian conditional probability model. It might help to read up on how it works.
  $endgroup$
  – Don Thousand
  Apr 6 at 16:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this makes sense right away. Define events $A_k = $ "the student knows $k$ questions", $B_k$ = "the student answers $k$ questions right" . You are looking for $mathbbP(A_1 cup dots cup A_7 | B_9)$ and you can use Bayes formulas to compute this.
  $endgroup$
  – Gâteau-Gallois
  Apr 6 at 16:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gâteau-Gallois am I wrong that if we use direct bayes formula, we will need to know $P(A_i)$ which is not given?
  $endgroup$
  – Mikhail Kuzin
  Apr 6 at 17:00
  

  
  
  
  

add a comment | 

0

$begingroup$

The problem forumaltes as:

Student takes test of 12 questions, each question is single-choice with 4 options. For each question he answers correctly if he knows the question, otherwise he guesses uniformly. Assume student answered correctly for exactly 9 questions. What is probability that he knew at least 7 questions?
My question is the following: does this problem statement make sense unless we have a priori probability distribution of how many questions does student know?

If it makes sense, how to solve it, and if it doesn't, how to show that and what could be the ways to complete the statement of this problem so that it would make sense?

probability-theory bayes-theorem

share|cite|improve this question

edited Apr 6 at 16:54

Bernard

124k741117

asked Apr 6 at 16:47

Mikhail KuzinMikhail Kuzin

304

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This question does make sense, assuming the Bayesian conditional probability model. It might help to read up on how it works.
  $endgroup$
  – Don Thousand
  Apr 6 at 16:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think this makes sense right away. Define events $A_k = $ "the student knows $k$ questions", $B_k$ = "the student answers $k$ questions right" . You are looking for $mathbbP(A_1 cup dots cup A_7 | B_9)$ and you can use Bayes formulas to compute this.
  $endgroup$
  – Gâteau-Gallois
  Apr 6 at 16:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gâteau-Gallois am I wrong that if we use direct bayes formula, we will need to know $P(A_i)$ which is not given?
  $endgroup$
  – Mikhail Kuzin
  Apr 6 at 17:00
  

  
  
  
  

add a comment | 

$begingroup$

The problem forumaltes as:

Student takes test of 12 questions, each question is single-choice with 4 options. For each question he answers correctly if he knows the question, otherwise he guesses uniformly. Assume student answered correctly for exactly 9 questions. What is probability that he knew at least 7 questions?
My question is the following: does this problem statement make sense unless we have a priori probability distribution of how many questions does student know?

If it makes sense, how to solve it, and if it doesn't, how to show that and what could be the ways to complete the statement of this problem so that it would make sense?

probability-theory bayes-theorem

share|cite|improve this question

edited Apr 6 at 16:54

Bernard

124k741117

asked Apr 6 at 16:47

Mikhail KuzinMikhail Kuzin

304

$endgroup$

share|cite|improve this question

edited Apr 6 at 16:54

Bernard

124k741117

asked Apr 6 at 16:47

Mikhail KuzinMikhail Kuzin

304

share|cite|improve this question

edited Apr 6 at 16:54

Bernard

124k741117

asked Apr 6 at 16:47

Mikhail KuzinMikhail Kuzin

304

edited Apr 6 at 16:54

Bernard

124k741117

edited Apr 6 at 16:54

Bernard

124k741117

asked Apr 6 at 16:47

Mikhail KuzinMikhail Kuzin

304

asked Apr 6 at 16:47

Mikhail KuzinMikhail Kuzin

304

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $k$ denote the number of questions the students genuinely knows how to answer. Clearly $kle 9$. Then the probability of scoring $9$ is $fracbinom12-k33^34^12-k$. Using Bayes's theorem, we can write $k$'s probability distribution conditional on a score $S=9$ as $$P(k=r|S=9)=fracbinom12-r33^34^12-rfracP(k=r)P(S=9).$$In particular,$$P(k=7|S=9)=fracbinom533^34^5fracP(k=7)P(S=9).$$But $P(k=7)$ is a Bayesian prior, and $P(S=9)$ is obtained from that prior. So if you want to get a specific numerical value for the probability, we need more information. For example, suppose the questions are independently known with probability $p$, so $$P(k=r)=binom12rp^r(1-p)^12-r,,P(S=9)=sum_r=0^9fracbinom12-r33^34^12-rP(k=r).$$Ultimately, $P(k=7|S=9)$ will depend on $p$. Bayesian statistics always starts from a prior; the posterior depends on the prior, but with less sensitivity to it given a large dataset.

share|cite|improve this answer

answered Apr 6 at 17:07

J.G.J.G.

33.1k23252

$endgroup$

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3177211%2fdoes-this-elementary-probability-theory-problem-make-sense%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Let $k$ denote the number of questions the students genuinely knows how to answer. Clearly $kle 9$. Then the probability of scoring $9$ is $fracbinom12-k33^34^12-k$. Using Bayes's theorem, we can write $k$'s probability distribution conditional on a score $S=9$ as $$P(k=r|S=9)=fracbinom12-r33^34^12-rfracP(k=r)P(S=9).$$In particular,$$P(k=7|S=9)=fracbinom533^34^5fracP(k=7)P(S=9).$$But $P(k=7)$ is a Bayesian prior, and $P(S=9)$ is obtained from that prior. So if you want to get a specific numerical value for the probability, we need more information. For example, suppose the questions are independently known with probability $p$, so $$P(k=r)=binom12rp^r(1-p)^12-r,,P(S=9)=sum_r=0^9fracbinom12-r33^34^12-rP(k=r).$$Ultimately, $P(k=7|S=9)$ will depend on $p$. Bayesian statistics always starts from a prior; the posterior depends on the prior, but with less sensitivity to it given a large dataset.

share|cite|improve this answer

answered Apr 6 at 17:07

J.G.J.G.

33.1k23252

$endgroup$

add a comment | 

1

$begingroup$

Let $k$ denote the number of questions the students genuinely knows how to answer. Clearly $kle 9$. Then the probability of scoring $9$ is $fracbinom12-k33^34^12-k$. Using Bayes's theorem, we can write $k$'s probability distribution conditional on a score $S=9$ as $$P(k=r|S=9)=fracbinom12-r33^34^12-rfracP(k=r)P(S=9).$$In particular,$$P(k=7|S=9)=fracbinom533^34^5fracP(k=7)P(S=9).$$But $P(k=7)$ is a Bayesian prior, and $P(S=9)$ is obtained from that prior. So if you want to get a specific numerical value for the probability, we need more information. For example, suppose the questions are independently known with probability $p$, so $$P(k=r)=binom12rp^r(1-p)^12-r,,P(S=9)=sum_r=0^9fracbinom12-r33^34^12-rP(k=r).$$Ultimately, $P(k=7|S=9)$ will depend on $p$. Bayesian statistics always starts from a prior; the posterior depends on the prior, but with less sensitivity to it given a large dataset.

share|cite|improve this answer

answered Apr 6 at 17:07

J.G.J.G.

33.1k23252

$endgroup$

add a comment | 

$begingroup$

Let $k$ denote the number of questions the students genuinely knows how to answer. Clearly $kle 9$. Then the probability of scoring $9$ is $fracbinom12-k33^34^12-k$. Using Bayes's theorem, we can write $k$'s probability distribution conditional on a score $S=9$ as $$P(k=r|S=9)=fracbinom12-r33^34^12-rfracP(k=r)P(S=9).$$In particular,$$P(k=7|S=9)=fracbinom533^34^5fracP(k=7)P(S=9).$$But $P(k=7)$ is a Bayesian prior, and $P(S=9)$ is obtained from that prior. So if you want to get a specific numerical value for the probability, we need more information. For example, suppose the questions are independently known with probability $p$, so $$P(k=r)=binom12rp^r(1-p)^12-r,,P(S=9)=sum_r=0^9fracbinom12-r33^34^12-rP(k=r).$$Ultimately, $P(k=7|S=9)$ will depend on $p$. Bayesian statistics always starts from a prior; the posterior depends on the prior, but with less sensitivity to it given a large dataset.

share|cite|improve this answer

answered Apr 6 at 17:07

J.G.J.G.

33.1k23252

$endgroup$

share|cite|improve this answer

answered Apr 6 at 17:07

J.G.J.G.

33.1k23252

answered Apr 6 at 17:07

J.G.J.G.

33.1k23252

answered Apr 6 at 17:07

J.G.J.G.

33.1k23252