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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
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Ramblin Wreck is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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♦
4 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♦
4 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♦
4 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♦
4 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♦
4 hours ago
add a comment |
1 Answer
1
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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
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$endgroup$
$begingroup$
(+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
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1 Answer
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$begingroup$
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
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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.
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$endgroup$
$begingroup$
(+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.
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answered 3 hours ago
dlnBdlnB
3264
answered 3 hours ago
dlnBdlnB
3264
3264
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dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
(+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.
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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♦
4 hours ago