Fdtxjr

Multi tool use

What will happen if my luggage gets delayed?

Does a difference of tense count as a difference of meaning in a minimal pair?

Which situations would cause a company to ground or recall a aircraft series?

Is it possible that a question has only two answers?

What would be the most expensive material to an intergalactic society?

Making a kiddush for a girl that has hard time finding shidduch

Proving a statement about real numbers

Having the player face themselves after the mid-game

Giving a career talk in my old university, how prominently should I tell students my salary?

How can I manipulate the output of Information?

Recommendation letter by significant other if you worked with them professionally?

Why couldn't the separatists legally leave the Republic?

Vocabulary for giving just numbers, not a full answer

Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?

Does "Until when" sound natural for native speakers?

Why does cron require MTA for logging?

What is the population of Romulus in the TNG era?

Can the alpha, lambda values of a glmnet object output determine whether ridge or Lasso?

How do we create new idioms and use them in a novel?

Do cubics always have one real root?

When a wind turbine does not produce enough electricity how does the power company compensate for the loss?

Do I really need to have a scientific explanation for my premise?

What is this diamond of every day?

After `ssh` without `-X` to a machine, is it possible to change `$DISPLAY` to make it work like `ssh -X`?

Probability of there being a bullet in the chamber

Probability Theory Question on Expected Value and Variance of Random VariableRussian roulette should a player pull the trigger or spin the cylinderColliding BulletsA probability question that I failed to answer in a job interviewProving russian roulette survival statisticsLoaded revolver puzzle.Russian Roulette variant, option to point awayCalculating probability and optimization - russian rouletteExpected value of Russian roulette at a firing range played on end with finite total shotsRussian Roulette Probability

1

$begingroup$

Suppose we have three $6$-chambered guns:

• The first has no bullets.




• The second has one bullet.




• The third has two bullets in consecutive chambers.

The cylinder advances automatically as the trigger is pulled. A man grabs a revolver at random, aims it at him, pulls the trigger $t$ times, and no shot is fired. He then aims at your head and pulls the trigger once. What is the probability you are shot?

probability conditional-probability word-problem

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The man "grabs a revolver at random" but what is the distribution? Uniform?
  $endgroup$
  – Rodrigo de Azevedo
  yesterday
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Suppose we have three $6$-chambered guns:

• The first has no bullets.




• The second has one bullet.




• The third has two bullets in consecutive chambers.

The cylinder advances automatically as the trigger is pulled. A man grabs a revolver at random, aims it at him, pulls the trigger $t$ times, and no shot is fired. He then aims at your head and pulls the trigger once. What is the probability you are shot?

probability conditional-probability word-problem

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The man "grabs a revolver at random" but what is the distribution? Uniform?
  $endgroup$
  – Rodrigo de Azevedo
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

Suppose we have three $6$-chambered guns:

• The first has no bullets.




• The second has one bullet.




• The third has two bullets in consecutive chambers.

The cylinder advances automatically as the trigger is pulled. A man grabs a revolver at random, aims it at him, pulls the trigger $t$ times, and no shot is fired. He then aims at your head and pulls the trigger once. What is the probability you are shot?

probability conditional-probability word-problem

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

$endgroup$

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

share|cite|improve this question

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

edited yesterday

Rodrigo de Azevedo

13k41960

edited yesterday

Rodrigo de Azevedo

13k41960

asked yesterday

KombatWombatKombatWombat

756

asked yesterday

KombatWombatKombatWombat

756

1 Answer
1

active

oldest

votes

2

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i'm struggling to understand your answer, could you possible show how I would obtain the answer for t=1 using your way please.
  $endgroup$
  – KombatWombat
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm more confused as to how you set up the problem, i.e how you calculate P(X=2) and probability (X=infinity). I understand for x = 1 is number of bullets(3) divided by number of chambers.
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how do you know that you should combine them making it out of 18 instead of multiplying each one out of 6 by 1/3?
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $t=1$ we find $P(X=2mid X>1)=P(X=2)/P(X>1)=2/(18-3)=2/15$. Note that there are $15$ empty chambers and exactly $2$ of them are followed by a chamber that contains a bullet. Concerning your last comment: it is not so much a matter of "knowing" but more a (nice) way to approach the problem. Instead of looking at $3$ guns we are looking at $18$ equiprobable chambers.
  $endgroup$
  – drhab
  yesterday
  
  

  
  
  
  

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%2f3140874%2fprobability-of-there-being-a-bullet-in-the-chamber%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i'm struggling to understand your answer, could you possible show how I would obtain the answer for t=1 using your way please.
  $endgroup$
  – KombatWombat
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm more confused as to how you set up the problem, i.e how you calculate P(X=2) and probability (X=infinity). I understand for x = 1 is number of bullets(3) divided by number of chambers.
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how do you know that you should combine them making it out of 18 instead of multiplying each one out of 6 by 1/3?
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $t=1$ we find $P(X=2mid X>1)=P(X=2)/P(X>1)=2/(18-3)=2/15$. Note that there are $15$ empty chambers and exactly $2$ of them are followed by a chamber that contains a bullet. Concerning your last comment: it is not so much a matter of "knowing" but more a (nice) way to approach the problem. Instead of looking at $3$ guns we are looking at $18$ equiprobable chambers.
  $endgroup$
  – drhab
  yesterday
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i'm struggling to understand your answer, could you possible show how I would obtain the answer for t=1 using your way please.
  $endgroup$
  – KombatWombat
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm more confused as to how you set up the problem, i.e how you calculate P(X=2) and probability (X=infinity). I understand for x = 1 is number of bullets(3) divided by number of chambers.
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, how do you know that you should combine them making it out of 18 instead of multiplying each one out of 6 by 1/3?
  $endgroup$
  – KombatWombat
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For $t=1$ we find $P(X=2mid X>1)=P(X=2)/P(X>1)=2/(18-3)=2/15$. Note that there are $15$ empty chambers and exactly $2$ of them are followed by a chamber that contains a bullet. Concerning your last comment: it is not so much a matter of "knowing" but more a (nice) way to approach the problem. Instead of looking at $3$ guns we are looking at $18$ equiprobable chambers.
  $endgroup$
  – drhab
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

Setup:

Let $X$ denote the number of shots needed for appearing a bullet if at random a revolver is picked and is shooted several times.

Then $X$ takes values in $left1,2,3,4,5,6,inftyright$

Note that there are $18$ chambers in total having equal probability
to be chosen for the first shot, and it is not really difficult to find that:

$Pleft(X=1right)=frac318$

$Pleft(X=2right)=Pleft(X=3right)=Pleft(X=4right)=Pleft(X=5right)=frac218$

$Pleft(X=6right)=frac118$

$Pleft(X=inftyright)=frac618$

Based on this you can find $P(X=t+1mid X>t)$.

Observe that this probability takes value $0$ if $t>5$ so the task can be completed by finding expressions for $tin0,1,2,3,4,5$.

I leave the rest to you.

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

edited yesterday

Rodrigo de Azevedo

13k41960

edited yesterday

Rodrigo de Azevedo

13k41960

answered yesterday

drhabdrhab

103k545136

answered yesterday

drhabdrhab

103k545136