Old race car problem/puzzleThe Monty Hall Problem - where does our intuition fail us?How to solve Chuck-a-Luck puzzleTwo envelope problem revisitedTicket selling probability puzzleModified German Tank problem involving string distance metricsProbability puzzle solved by R simulation; answer close but not exactly. Bug in my code?
What has been your most complicated TikZ drawing?
How to explain that I do not want to visit a country due to personal safety concern?
Why is stat::st_size 0 for devices but at the same time lseek defines the device size correctly?
Font with correct density?
Russian cases: A few examples, I'm really confused
RegionDifference for Cylinder and Cuboid
Employee lack of ownership
How to deal with taxi scam when on vacation?
My adviser wants to be the first author
Does this AnyDice function accurately calculate the number of ogres you make unconcious with three 4th-level castings of Sleep?
Is having access to past exams cheating and, if yes, could it be proven just by a good grade?
In-house repeater?
How Did the Space Junk Stay in Orbit in Wall-E?
Did CPM support custom hardware using device drivers?
What is IP squat space
Where is the 1/8 CR apprentice in Volo's Guide to Monsters?
Is it normal that my co-workers at a fitness company criticize my food choices?
Make a transparent 448*448 image
The use of "touch" and "touch on" in context
Rejected in 4th interview round citing insufficient years of experience
What is the greatest age difference between a married couple in Tanach?
Counting certain elements in lists
Meaning of "SEVERA INDEOVI VAS" from 3rd Century slab
Theorems like the Lovász Local Lemma?
Old race car problem/puzzle
The Monty Hall Problem - where does our intuition fail us?How to solve Chuck-a-Luck puzzleTwo envelope problem revisitedTicket selling probability puzzleModified German Tank problem involving string distance metricsProbability puzzle solved by R simulation; answer close but not exactly. Bug in my code?
$begingroup$
This was in an old (1935) "brain teaser" book, and I can't figure it out how to solve it!
There's a car race during which the cars experience 4 different types of car trouble, e.g. flat tire, blown motor, etc. I can't recall exactly what they are let's say they are A, B, C, and D.
Also it is stated that: 95% of the cars experience trouble A, 85% of the cars experience trouble B, 75% of the cars experience trouble C, and 65% of the cars experience trouble D
Then the question that is asked is, what is the least percentage of cars which must have experienced all 4 car troubles?
Anybody have any ideas? Thanks!
puzzle
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This was in an old (1935) "brain teaser" book, and I can't figure it out how to solve it!
There's a car race during which the cars experience 4 different types of car trouble, e.g. flat tire, blown motor, etc. I can't recall exactly what they are let's say they are A, B, C, and D.
Also it is stated that: 95% of the cars experience trouble A, 85% of the cars experience trouble B, 75% of the cars experience trouble C, and 65% of the cars experience trouble D
Then the question that is asked is, what is the least percentage of cars which must have experienced all 4 car troubles?
Anybody have any ideas? Thanks!
puzzle
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Hint: when you rephrase the information in terms of percentages of cars that did not experience trouble, the answer might become obvious.
$endgroup$
– whuber♦
3 hours ago
add a comment |
$begingroup$
This was in an old (1935) "brain teaser" book, and I can't figure it out how to solve it!
There's a car race during which the cars experience 4 different types of car trouble, e.g. flat tire, blown motor, etc. I can't recall exactly what they are let's say they are A, B, C, and D.
Also it is stated that: 95% of the cars experience trouble A, 85% of the cars experience trouble B, 75% of the cars experience trouble C, and 65% of the cars experience trouble D
Then the question that is asked is, what is the least percentage of cars which must have experienced all 4 car troubles?
Anybody have any ideas? Thanks!
puzzle
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This was in an old (1935) "brain teaser" book, and I can't figure it out how to solve it!
There's a car race during which the cars experience 4 different types of car trouble, e.g. flat tire, blown motor, etc. I can't recall exactly what they are let's say they are A, B, C, and D.
Also it is stated that: 95% of the cars experience trouble A, 85% of the cars experience trouble B, 75% of the cars experience trouble C, and 65% of the cars experience trouble D
Then the question that is asked is, what is the least percentage of cars which must have experienced all 4 car troubles?
Anybody have any ideas? Thanks!
puzzle
puzzle
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Ramblin Wreck
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
asked 4 hours ago
Ramblin WreckRamblin Wreck
63
63
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Hint: when you rephrase the information in terms of percentages of cars that did not experience trouble, the answer might become obvious.
$endgroup$
– whuber♦
3 hours ago
add a comment |
1
$begingroup$
Hint: when you rephrase the information in terms of percentages of cars that did not experience trouble, the answer might become obvious.
$endgroup$
– whuber♦
3 hours ago
1
1
$begingroup$
Hint: when you rephrase the information in terms of percentages of cars that did not experience trouble, the answer might become obvious.
$endgroup$
– whuber♦
3 hours ago
$begingroup$
Hint: when you rephrase the information in terms of percentages of cars that did not experience trouble, the answer might become obvious.
$endgroup$
– whuber♦
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-P((A cap B cap C cap D)^c)=1-P(A^c cup B^c cup C^c cup D^c).$$
Notice the upper bound on $P(A^c cup B^c cup C^c cup D^c)$ (and therefore the lower bound on $P(A cap B cap C cap D)$) is given by the case where the four sets $A^c, B^c, C^c,$ and $D^c$ are mutually exclusive (it helps to draw a picture to better understand this). In this case
$$P(A^c cup B^c cup C^c cup D^c)=P(A^c)+P(B^c)+P(C^c)+P(D^C)=.35+.25+.15+.05=0.8,$$
since $P(A^c)=.35, P(B^c)=.25, P(C^c)=.15,$ and $P(D^c)=.05$. Thus, the upper bound on the probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-0.8=0.2.$$
The reason we cannot exactly calculate the probability $P(A cap B cap C cap D)$ is that we do not have information about the probability of set intersections. We can, however, find an upper bound, which is given by the case where the sets do not overlap.
answered 3 hours ago
dlnBdlnB
3264
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
(+1) but note that it should say "(and therefore the lower bound on $P(A cap B cap C cap D)$) "
$endgroup$
– dnqxt
3 hours ago
$begingroup$
Yes, thank you. Corrected.
$endgroup$
– dlnB
3 hours ago
$begingroup$
Thank you for the clear explanation! Drawing a picture of the mutually exclusive sets (for a small number of cars) helped me get it. Also, 20% is what the book had for the answer too.
$endgroup$
– Ramblin Wreck
3 hours ago
$begingroup$
Pictures usually help with the intuition. Glad I could help. If you are satisfied, please accept the answer.
$endgroup$
– dlnB
2 hours ago
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: "65"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Ramblin Wreck is a new contributor. Be nice, and check out our Code of Conduct.
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%2fstats.stackexchange.com%2fquestions%2f397568%2fold-race-car-problem-puzzle%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
$begingroup$
The probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-P((A cap B cap C cap D)^c)=1-P(A^c cup B^c cup C^c cup D^c).$$
Notice the upper bound on $P(A^c cup B^c cup C^c cup D^c)$ (and therefore the lower bound on $P(A cap B cap C cap D)$) is given by the case where the four sets $A^c, B^c, C^c,$ and $D^c$ are mutually exclusive (it helps to draw a picture to better understand this). In this case
$$P(A^c cup B^c cup C^c cup D^c)=P(A^c)+P(B^c)+P(C^c)+P(D^C)=.35+.25+.15+.05=0.8,$$
since $P(A^c)=.35, P(B^c)=.25, P(C^c)=.15,$ and $P(D^c)=.05$. Thus, the upper bound on the probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-0.8=0.2.$$
The reason we cannot exactly calculate the probability $P(A cap B cap C cap D)$ is that we do not have information about the probability of set intersections. We can, however, find an upper bound, which is given by the case where the sets do not overlap.
answered 3 hours ago
dlnBdlnB
3264
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
(+1) but note that it should say "(and therefore the lower bound on $P(A cap B cap C cap D)$) "
$endgroup$
– dnqxt
3 hours ago
$begingroup$
Yes, thank you. Corrected.
$endgroup$
– dlnB
3 hours ago
$begingroup$
Thank you for the clear explanation! Drawing a picture of the mutually exclusive sets (for a small number of cars) helped me get it. Also, 20% is what the book had for the answer too.
$endgroup$
– Ramblin Wreck
3 hours ago
$begingroup$
Pictures usually help with the intuition. Glad I could help. If you are satisfied, please accept the answer.
$endgroup$
– dlnB
2 hours ago
add a comment |
$begingroup$
The probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-P((A cap B cap C cap D)^c)=1-P(A^c cup B^c cup C^c cup D^c).$$
Notice the upper bound on $P(A^c cup B^c cup C^c cup D^c)$ (and therefore the lower bound on $P(A cap B cap C cap D)$) is given by the case where the four sets $A^c, B^c, C^c,$ and $D^c$ are mutually exclusive (it helps to draw a picture to better understand this). In this case
$$P(A^c cup B^c cup C^c cup D^c)=P(A^c)+P(B^c)+P(C^c)+P(D^C)=.35+.25+.15+.05=0.8,$$
since $P(A^c)=.35, P(B^c)=.25, P(C^c)=.15,$ and $P(D^c)=.05$. Thus, the upper bound on the probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-0.8=0.2.$$
The reason we cannot exactly calculate the probability $P(A cap B cap C cap D)$ is that we do not have information about the probability of set intersections. We can, however, find an upper bound, which is given by the case where the sets do not overlap.
answered 3 hours ago
dlnBdlnB
3264
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
(+1) but note that it should say "(and therefore the lower bound on $P(A cap B cap C cap D)$) "
$endgroup$
– dnqxt
3 hours ago
$begingroup$
Yes, thank you. Corrected.
$endgroup$
– dlnB
3 hours ago
$begingroup$
Thank you for the clear explanation! Drawing a picture of the mutually exclusive sets (for a small number of cars) helped me get it. Also, 20% is what the book had for the answer too.
$endgroup$
– Ramblin Wreck
3 hours ago
$begingroup$
Pictures usually help with the intuition. Glad I could help. If you are satisfied, please accept the answer.
$endgroup$
– dlnB
2 hours ago
add a comment |
$begingroup$
The probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-P((A cap B cap C cap D)^c)=1-P(A^c cup B^c cup C^c cup D^c).$$
Notice the upper bound on $P(A^c cup B^c cup C^c cup D^c)$ (and therefore the lower bound on $P(A cap B cap C cap D)$) is given by the case where the four sets $A^c, B^c, C^c,$ and $D^c$ are mutually exclusive (it helps to draw a picture to better understand this). In this case
$$P(A^c cup B^c cup C^c cup D^c)=P(A^c)+P(B^c)+P(C^c)+P(D^C)=.35+.25+.15+.05=0.8,$$
since $P(A^c)=.35, P(B^c)=.25, P(C^c)=.15,$ and $P(D^c)=.05$. Thus, the upper bound on the probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-0.8=0.2.$$
The reason we cannot exactly calculate the probability $P(A cap B cap C cap D)$ is that we do not have information about the probability of set intersections. We can, however, find an upper bound, which is given by the case where the sets do not overlap.
answered 3 hours ago
dlnBdlnB
3264
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-P((A cap B cap C cap D)^c)=1-P(A^c cup B^c cup C^c cup D^c).$$
Notice the upper bound on $P(A^c cup B^c cup C^c cup D^c)$ (and therefore the lower bound on $P(A cap B cap C cap D)$) is given by the case where the four sets $A^c, B^c, C^c,$ and $D^c$ are mutually exclusive (it helps to draw a picture to better understand this). In this case
$$P(A^c cup B^c cup C^c cup D^c)=P(A^c)+P(B^c)+P(C^c)+P(D^C)=.35+.25+.15+.05=0.8,$$
since $P(A^c)=.35, P(B^c)=.25, P(C^c)=.15,$ and $P(D^c)=.05$. Thus, the upper bound on the probability of having car troubles A, B, C, and D is given by
$$P(A cap B cap C cap D)=1-0.8=0.2.$$
The reason we cannot exactly calculate the probability $P(A cap B cap C cap D)$ is that we do not have information about the probability of set intersections. We can, however, find an upper bound, which is given by the case where the sets do not overlap.
answered 3 hours ago
dlnBdlnB
3264
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
answered 3 hours ago
dlnBdlnB
3264
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
dlnBdlnB
3264
answered 3 hours ago
dlnBdlnB
3264
3264
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
(+1) but note that it should say "(and therefore the lower bound on $P(A cap B cap C cap D)$) "
$endgroup$
– dnqxt
3 hours ago
$begingroup$
Yes, thank you. Corrected.
$endgroup$
– dlnB
3 hours ago
$begingroup$
Thank you for the clear explanation! Drawing a picture of the mutually exclusive sets (for a small number of cars) helped me get it. Also, 20% is what the book had for the answer too.
$endgroup$
– Ramblin Wreck
3 hours ago
$begingroup$
Pictures usually help with the intuition. Glad I could help. If you are satisfied, please accept the answer.
$endgroup$
– dlnB
2 hours ago
add a comment |
$begingroup$
(+1) but note that it should say "(and therefore the lower bound on $P(A cap B cap C cap D)$) "
$endgroup$
– dnqxt
3 hours ago
$begingroup$
Yes, thank you. Corrected.
$endgroup$
– dlnB
3 hours ago
$begingroup$
Thank you for the clear explanation! Drawing a picture of the mutually exclusive sets (for a small number of cars) helped me get it. Also, 20% is what the book had for the answer too.
$endgroup$
– Ramblin Wreck
3 hours ago
$begingroup$
Pictures usually help with the intuition. Glad I could help. If you are satisfied, please accept the answer.
$endgroup$
– dlnB
2 hours ago
$begingroup$
(+1) but note that it should say "(and therefore the lower bound on $P(A cap B cap C cap D)$) "
$endgroup$
– dnqxt
3 hours ago
$begingroup$
(+1) but note that it should say "(and therefore the lower bound on $P(A cap B cap C cap D)$) "
$endgroup$
– dnqxt
3 hours ago
$begingroup$
Yes, thank you. Corrected.
$endgroup$
– dlnB
3 hours ago
$begingroup$
Yes, thank you. Corrected.
$endgroup$
– dlnB
3 hours ago
$begingroup$
Thank you for the clear explanation! Drawing a picture of the mutually exclusive sets (for a small number of cars) helped me get it. Also, 20% is what the book had for the answer too.
$endgroup$
– Ramblin Wreck
3 hours ago
$begingroup$
Thank you for the clear explanation! Drawing a picture of the mutually exclusive sets (for a small number of cars) helped me get it. Also, 20% is what the book had for the answer too.
$endgroup$
– Ramblin Wreck
3 hours ago
$begingroup$
Pictures usually help with the intuition. Glad I could help. If you are satisfied, please accept the answer.
$endgroup$
– dlnB
2 hours ago
$begingroup$
Pictures usually help with the intuition. Glad I could help. If you are satisfied, please accept the answer.
$endgroup$
– dlnB
2 hours ago
add a comment |
Ramblin Wreck is a new contributor. Be nice, and check out our Code of Conduct.
Ramblin Wreck is a new contributor. Be nice, and check out our Code of Conduct.
Ramblin Wreck is a new contributor. Be nice, and check out our Code of Conduct.
Ramblin Wreck is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Cross Validated!
- 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%2fstats.stackexchange.com%2fquestions%2f397568%2fold-race-car-problem-puzzle%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
$begingroup$
Hint: when you rephrase the information in terms of percentages of cars that did not experience trouble, the answer might become obvious.
$endgroup$
– whuber♦
3 hours ago