Cube and unit cubes
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Consider a $3times 3times 3$ cube consisting of smaller $1times 1times 1$ unit cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and pick a random unit cube, look at only one face and see it is black, without looking at the other faces. What is the probability the unit cube we picked is one of the 8 corner cubes?
This one seems simple to me but I am not sure I am right:
The big cube consists of 27 unit cubes of which one only, the middle one, does not have any painted face. All the others (26) have at least one face painted and 8 have 3 faces painted.
Thus the requested probability is $frac826$—is this so?
probability
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
$endgroup$
add a comment |
$begingroup$
Consider a $3times 3times 3$ cube consisting of smaller $1times 1times 1$ unit cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and pick a random unit cube, look at only one face and see it is black, without looking at the other faces. What is the probability the unit cube we picked is one of the 8 corner cubes?
This one seems simple to me but I am not sure I am right:
The big cube consists of 27 unit cubes of which one only, the middle one, does not have any painted face. All the others (26) have at least one face painted and 8 have 3 faces painted.
Thus the requested probability is $frac826$—is this so?
probability
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
$endgroup$
1
$begingroup$
There are already good answers, but the problem you solved with answer $8/26$ is a different one: Say I randomly select a unit cube and look at all of its faces. You ask me if at least one face on that cube is black, and I answer "yes". Given that, what is the probability the unit cube I picked is a corner cube? The major difference is knowing something (though not everything) about all the cube's faces, or knowing about just one face of a cube.
$endgroup$
– aschepler
Feb 16 at 19:55
add a comment |
$begingroup$
Consider a $3times 3times 3$ cube consisting of smaller $1times 1times 1$ unit cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and pick a random unit cube, look at only one face and see it is black, without looking at the other faces. What is the probability the unit cube we picked is one of the 8 corner cubes?
This one seems simple to me but I am not sure I am right:
The big cube consists of 27 unit cubes of which one only, the middle one, does not have any painted face. All the others (26) have at least one face painted and 8 have 3 faces painted.
Thus the requested probability is $frac826$—is this so?
probability
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
$endgroup$
Consider a $3times 3times 3$ cube consisting of smaller $1times 1times 1$ unit cubes. The big cube is painted black on the outside. Suppose we disassemble the cube and pick a random unit cube, look at only one face and see it is black, without looking at the other faces. What is the probability the unit cube we picked is one of the 8 corner cubes?
This one seems simple to me but I am not sure I am right:
The big cube consists of 27 unit cubes of which one only, the middle one, does not have any painted face. All the others (26) have at least one face painted and 8 have 3 faces painted.
Thus the requested probability is $frac826$—is this so?
probability
probability
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
edited Feb 16 at 16:59
Xander Henderson
14.9k103555
14.9k103555
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
asked Feb 16 at 16:56
Sal.CognatoSal.Cognato
33019
33019
1
$begingroup$
There are already good answers, but the problem you solved with answer $8/26$ is a different one: Say I randomly select a unit cube and look at all of its faces. You ask me if at least one face on that cube is black, and I answer "yes". Given that, what is the probability the unit cube I picked is a corner cube? The major difference is knowing something (though not everything) about all the cube's faces, or knowing about just one face of a cube.
$endgroup$
– aschepler
Feb 16 at 19:55
add a comment |
1
$begingroup$
There are already good answers, but the problem you solved with answer $8/26$ is a different one: Say I randomly select a unit cube and look at all of its faces. You ask me if at least one face on that cube is black, and I answer "yes". Given that, what is the probability the unit cube I picked is a corner cube? The major difference is knowing something (though not everything) about all the cube's faces, or knowing about just one face of a cube.
$endgroup$
– aschepler
Feb 16 at 19:55
1
1
$begingroup$
There are already good answers, but the problem you solved with answer $8/26$ is a different one: Say I randomly select a unit cube and look at all of its faces. You ask me if at least one face on that cube is black, and I answer "yes". Given that, what is the probability the unit cube I picked is a corner cube? The major difference is knowing something (though not everything) about all the cube's faces, or knowing about just one face of a cube.
$endgroup$
– aschepler
Feb 16 at 19:55
$begingroup$
There are already good answers, but the problem you solved with answer $8/26$ is a different one: Say I randomly select a unit cube and look at all of its faces. You ask me if at least one face on that cube is black, and I answer "yes". Given that, what is the probability the unit cube I picked is a corner cube? The major difference is knowing something (though not everything) about all the cube's faces, or knowing about just one face of a cube.
$endgroup$
– aschepler
Feb 16 at 19:55
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.
Here is a possible line of reasoning: when you paint the $3times 3times 3$ cube, you paint a total of
$$ 9 cdot 6 = 54 $$
faces (each face of the large cube consists of a $3times 3$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).
On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of
$$ 3 cdot 8 = 24 $$
painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e.
$$
P(textthe painted face is from a corner)
= fractextnumber of successful outcomestexttotal number of outcomes
= frac2454
= frac49. $$
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
$endgroup$
add a comment |
$begingroup$
No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube?
The simple solution is that we have selected a random black face. There are $54$ black faces, $24$ of which are on corner cubes, so the chance we have a corner cube is $frac 2454$
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair!
$endgroup$
– Sal.Cognato
Feb 16 at 17:18
add a comment |
$begingroup$
You can also use Bayes' theorem. The prior probability of choosing a corner cube is $8 over 27$. The probability of a corner cube showing a black face is $1 over 2$. The overall probability of a black face is $54 over 162=1/3$, so
$P(rm corner |rm black)=1/2 over1/3 times 8/27 =4/9$
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.
Here is a possible line of reasoning: when you paint the $3times 3times 3$ cube, you paint a total of
$$ 9 cdot 6 = 54 $$
faces (each face of the large cube consists of a $3times 3$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).
On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of
$$ 3 cdot 8 = 24 $$
painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e.
$$
P(textthe painted face is from a corner)
= fractextnumber of successful outcomestexttotal number of outcomes
= frac2454
= frac49. $$
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
$endgroup$
add a comment |
$begingroup$
Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.
Here is a possible line of reasoning: when you paint the $3times 3times 3$ cube, you paint a total of
$$ 9 cdot 6 = 54 $$
faces (each face of the large cube consists of a $3times 3$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).
On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of
$$ 3 cdot 8 = 24 $$
painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e.
$$
P(textthe painted face is from a corner)
= fractextnumber of successful outcomestexttotal number of outcomes
= frac2454
= frac49. $$
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
$endgroup$
add a comment |
$begingroup$
Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.
Here is a possible line of reasoning: when you paint the $3times 3times 3$ cube, you paint a total of
$$ 9 cdot 6 = 54 $$
faces (each face of the large cube consists of a $3times 3$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).
On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of
$$ 3 cdot 8 = 24 $$
painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e.
$$
P(textthe painted face is from a corner)
= fractextnumber of successful outcomestexttotal number of outcomes
= frac2454
= frac49. $$
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
$endgroup$
Your first step is reasonable: if you see a painted face, then you know that the cube you have cannot possibly be the center cube. However, you are not "just as likely" to have a corner cube as a side cube. You need to be more careful in thinking about your sample space.
Here is a possible line of reasoning: when you paint the $3times 3times 3$ cube, you paint a total of
$$ 9 cdot 6 = 54 $$
faces (each face of the large cube consists of a $3times 3$ arrangement of faces of the smaller cubes; the large cube has six faces). This represents the total space of possible outcomes (given that you know you have selected a painted face—a priori, it might have been possible to selected a non-painted face, but we know this didn't happen).
On the other hand, there are 8 corner cubes, each of which as three painted faces, for a total of
$$ 3 cdot 8 = 24 $$
painted faces on corner cubes. Picking a painted face from a corner is a "success" in this experiment. Therefore the probability that you have selected a corner cube is the ratio of successes to the total number of possible outcomes, i.e.
$$
P(textthe painted face is from a corner)
= fractextnumber of successful outcomestexttotal number of outcomes
= frac2454
= frac49. $$
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
answered Feb 16 at 17:09
Xander HendersonXander Henderson
14.9k103555
14.9k103555
add a comment |
add a comment |
$begingroup$
No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube?
The simple solution is that we have selected a random black face. There are $54$ black faces, $24$ of which are on corner cubes, so the chance we have a corner cube is $frac 2454$
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair!
$endgroup$
– Sal.Cognato
Feb 16 at 17:18
add a comment |
$begingroup$
No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube?
The simple solution is that we have selected a random black face. There are $54$ black faces, $24$ of which are on corner cubes, so the chance we have a corner cube is $frac 2454$
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
$endgroup$
$begingroup$
Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair!
$endgroup$
– Sal.Cognato
Feb 16 at 17:18
add a comment |
$begingroup$
No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube?
The simple solution is that we have selected a random black face. There are $54$ black faces, $24$ of which are on corner cubes, so the chance we have a corner cube is $frac 2454$
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
$endgroup$
No, because you have picked a random face of the cube to look at and seen it is black. That makes it more likely that you have a corner cube than one of the other ones with some painted faces. Maybe a more careful statement of the problem is we choose a random cube and a random face of that cube. Given that the face we choose is black, what is the chance the cube is a corner cube?
The simple solution is that we have selected a random black face. There are $54$ black faces, $24$ of which are on corner cubes, so the chance we have a corner cube is $frac 2454$
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
answered Feb 16 at 17:08
Ross MillikanRoss Millikan
299k24200374
299k24200374
$begingroup$
Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair!
$endgroup$
– Sal.Cognato
Feb 16 at 17:18
add a comment |
$begingroup$
Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair!
$endgroup$
– Sal.Cognato
Feb 16 at 17:18
$begingroup$
Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair!
$endgroup$
– Sal.Cognato
Feb 16 at 17:18
$begingroup$
Thank you very much sirs! I gave acceptance to the one I see first. Hope I was not unfair!
$endgroup$
– Sal.Cognato
Feb 16 at 17:18
add a comment |
$begingroup$
You can also use Bayes' theorem. The prior probability of choosing a corner cube is $8 over 27$. The probability of a corner cube showing a black face is $1 over 2$. The overall probability of a black face is $54 over 162=1/3$, so
$P(rm corner |rm black)=1/2 over1/3 times 8/27 =4/9$
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
$endgroup$
add a comment |
$begingroup$
You can also use Bayes' theorem. The prior probability of choosing a corner cube is $8 over 27$. The probability of a corner cube showing a black face is $1 over 2$. The overall probability of a black face is $54 over 162=1/3$, so
$P(rm corner |rm black)=1/2 over1/3 times 8/27 =4/9$
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
$endgroup$
add a comment |
$begingroup$
You can also use Bayes' theorem. The prior probability of choosing a corner cube is $8 over 27$. The probability of a corner cube showing a black face is $1 over 2$. The overall probability of a black face is $54 over 162=1/3$, so
$P(rm corner |rm black)=1/2 over1/3 times 8/27 =4/9$
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
$endgroup$
You can also use Bayes' theorem. The prior probability of choosing a corner cube is $8 over 27$. The probability of a corner cube showing a black face is $1 over 2$. The overall probability of a black face is $54 over 162=1/3$, so
$P(rm corner |rm black)=1/2 over1/3 times 8/27 =4/9$
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
answered Feb 16 at 23:41
RogerJBarlowRogerJBarlow
2663
2663
add a comment |
add a comment |
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$begingroup$
There are already good answers, but the problem you solved with answer $8/26$ is a different one: Say I randomly select a unit cube and look at all of its faces. You ask me if at least one face on that cube is black, and I answer "yes". Given that, what is the probability the unit cube I picked is a corner cube? The major difference is knowing something (though not everything) about all the cube's faces, or knowing about just one face of a cube.
$endgroup$
– aschepler
Feb 16 at 19:55