The Pebbles Quiz
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We are in a TV game. We have 8 same looking boxes, each of which has 2 pebbles. Each pebble could be either precious or not precious. We choose a box and the host, without looking inside the box, pulls out a pebble. It comes out to be precious. The TV host then declares that we have exactly 50% chance that the 2nd pebble inside the box is precious, too.
If we know that, in the beginning of the game, the count of precious pebbles is no less than the count of not precious pebbles, which of the statements MUST be true?
We assume that the TV host knows the pebbles distribution in the boxes - in what amount of boxes the precious pebbles are 0,1 or 2. The TV host, however, doesn't know the kind of pebbles in each specific box, as he himself doesn't differ the boxes from one another.
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
None of the above.
mathematics logical-deduction
edited 12 hours ago
Brandon_J
2,415232
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
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add a comment |
$begingroup$
We are in a TV game. We have 8 same looking boxes, each of which has 2 pebbles. Each pebble could be either precious or not precious. We choose a box and the host, without looking inside the box, pulls out a pebble. It comes out to be precious. The TV host then declares that we have exactly 50% chance that the 2nd pebble inside the box is precious, too.
If we know that, in the beginning of the game, the count of precious pebbles is no less than the count of not precious pebbles, which of the statements MUST be true?
We assume that the TV host knows the pebbles distribution in the boxes - in what amount of boxes the precious pebbles are 0,1 or 2. The TV host, however, doesn't know the kind of pebbles in each specific box, as he himself doesn't differ the boxes from one another.
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
None of the above.
mathematics logical-deduction
edited 12 hours ago
Brandon_J
2,415232
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
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$endgroup$
$begingroup$
This is somewhat related to the somewhat annoying but extremely fascinating Monty Hall problem.
$endgroup$
– Brandon_J
7 hours ago
add a comment |
$begingroup$
We are in a TV game. We have 8 same looking boxes, each of which has 2 pebbles. Each pebble could be either precious or not precious. We choose a box and the host, without looking inside the box, pulls out a pebble. It comes out to be precious. The TV host then declares that we have exactly 50% chance that the 2nd pebble inside the box is precious, too.
If we know that, in the beginning of the game, the count of precious pebbles is no less than the count of not precious pebbles, which of the statements MUST be true?
We assume that the TV host knows the pebbles distribution in the boxes - in what amount of boxes the precious pebbles are 0,1 or 2. The TV host, however, doesn't know the kind of pebbles in each specific box, as he himself doesn't differ the boxes from one another.
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
None of the above.
mathematics logical-deduction
edited 12 hours ago
Brandon_J
2,415232
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
New contributor
Kradec na kysmet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We are in a TV game. We have 8 same looking boxes, each of which has 2 pebbles. Each pebble could be either precious or not precious. We choose a box and the host, without looking inside the box, pulls out a pebble. It comes out to be precious. The TV host then declares that we have exactly 50% chance that the 2nd pebble inside the box is precious, too.
If we know that, in the beginning of the game, the count of precious pebbles is no less than the count of not precious pebbles, which of the statements MUST be true?
We assume that the TV host knows the pebbles distribution in the boxes - in what amount of boxes the precious pebbles are 0,1 or 2. The TV host, however, doesn't know the kind of pebbles in each specific box, as he himself doesn't differ the boxes from one another.
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
None of the above.
mathematics logical-deduction
mathematics logical-deduction
edited 12 hours ago
Brandon_J
2,415232
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
New contributor
Kradec na kysmet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 12 hours ago
Brandon_J
2,415232
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
New contributor
Kradec na kysmet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 12 hours ago
Brandon_J
2,415232
edited 12 hours ago
Brandon_J
2,415232
edited 12 hours ago
Brandon_J
2,415232
2,415232
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
New contributor
Kradec na kysmet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
asked 18 hours ago
Kradec na kysmetKradec na kysmet
143
143
New contributor
Kradec na kysmet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kradec na kysmet is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
$begingroup$
This is somewhat related to the somewhat annoying but extremely fascinating Monty Hall problem.
$endgroup$
– Brandon_J
7 hours ago
add a comment |
$begingroup$
This is somewhat related to the somewhat annoying but extremely fascinating Monty Hall problem.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
This is somewhat related to the somewhat annoying but extremely fascinating Monty Hall problem.
$endgroup$
– Brandon_J
7 hours ago
$begingroup$
This is somewhat related to the somewhat annoying but extremely fascinating Monty Hall problem.
$endgroup$
– Brandon_J
7 hours ago
add a comment |
3 Answers
3
active
oldest
votes
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The answer is:
4, 5, 6, 7 must be true, while 1, 2, 3 are wrong.
Let us denote by $a, b, c$ the amount of boxes of each kind as follows:
PP PN NNa b c
First we know that $a+b+c=8$.
Now we use the $50%$ chance given by the host:
We apply Bayes' formula, where the event PP is "our box contains two precious pebbles" and P stands for "we pick a pebble from our box and it turns out to be precious".
We have $$0.5=Bbb P(P!P|P)=frac {Bbb P(P!P)}{Bbb P(P)}$$ $$0.5=Bbb P(P!P|P)=frac {a/8}{a/8+b/8cdot1/2}$$
So finally:
$$b=2a$$
Now the host also said that there are at least as many P's as N's. Therefore:
$$2a+bgeq 2c+b$$
which, given $a+b+c=8$ and $b=2a$, translates to
$$ageq 2$$
Partial conclusion
There is one configuration meeting these conditions, namely
PP PN NN2 4 2
and there is no other possibility, because $a=3$ would imply $b=6$ and $a+b$ would already be more than $8$.
We see that $4, 5, 6, 7$ are all true.
Moving on to questions 1, 2, 3:
If we have a PN box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 37cdot frac 12}_{PN}+underbrace{frac 27}_{PP}=0.5$$
If we have a PP box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 47cdot frac 12}_{PN}+underbrace{frac 17}_{PP}=frac 37$$
In the first case, the probability stays the same, while in the second it decreases. Therefore 1, 2, 3 are all wrong: it depends.
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
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Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host
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– Kradec na kysmet
17 hours ago
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@Kradecnakysmet is that an admission that statement 8 is false?
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– Weather Vane
17 hours ago
1
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You have a contradiction in your answer. You've calculated the probability of having a PP box as 2/3. If that were truly the case, then the chance that the second pebble in that box is precious is also 2/3. There's no other way around it. So either the host is lying, the calculated probability is wrong, or your original assumption of an equal number of PP and PN boxes is incorrect.
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– Amorydai
13 hours ago
1
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Also, since I deleted my answer, here's our chat discussion, Arnaud.
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– Brandon_J
8 hours ago
1
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Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes".
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– ppgdev
7 hours ago
|
show 17 more comments
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The selection procedure ensures that each of the 16 stones has an equal chance of being picked by the host.
Once his pick is revealed to be precious, it is still equally likely to be any one of the precious stones in the boxes. For the host's statement to be true, there must be the same number of precious stones with a precious stone partner as there are with a non-precious partner.
This means that for every box with two precious stones (which obviously both have a precious partner) there must be two boxes with a precious stone partnered with a non-precious stone.
This leaves two possibilities:
1xPP, 2xPN, 5xNN
2xPP, 4xPN, 2xNN
We are given that the number of precious stones is at least as many as the non-precious ones, so we must be in the second case:
2xPP, 4xPN, 2xNN
The other boxes have 6, or 7 precious stones in them, depending on whether the current box is PP or PN. You do not know which of these is the case, though you do know they are equally probable because the host said so. There are therefore on average 6.5 precious stones in the other boxes, and when switching boxes you have a probability of 6.5/14, or less than 50% to pick a precious stone next. Therefore statements 1 and 2 are false, 3 is true.
Statements 4-7 are easily seen to be true, and 8 and 9 are false.
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
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2
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2xPP + 4xPN doesn't make it 50% chance
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– Kradec na kysmet
16 hours ago
1
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@Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance.
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– Amorydai
13 hours ago
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@Kradecnakysmet yes it does. See my answer for another way to look at it.
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– Arnaud Mortier
10 hours ago
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@Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer.
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– Arnaud Mortier
10 hours ago
add a comment |
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There are two arrangements that satify the given facts:
A. 4 boxes with two precious pebbles, 4 boxes with one of each
B. 3 boxes with two precious pebbles, 3 with one of each, 2 with two non-precious
Because
There are at least 8 precious pebbles
Half of the boxes with a precious pebble contain another, half do not.
So for each statement:
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is higher.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
FALSE: in case A it is either more or less.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is lower.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
FALSE: in both solutions.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
FALSE: this is only true in one solution.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
FALSE: there are more precious pebbles in both solutions.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
FALSE: in both solutions.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
FALSE: there are two arrangements that satisfy the condition.
None of the above.
TRUE:
So my answer is
Statement 9 is the only true statement.
answered 18 hours ago
Weather VaneWeather Vane
1,55219
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For it to be true it has to be true for all conditions
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– Kradec na kysmet
16 hours ago
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You have to make sure it's true for all conditions
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– Kradec na kysmet
15 hours ago
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There's more than 1 condition, check it carefully
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– Kradec na kysmet
15 hours ago
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Recheck your answers, I see lots of mistakes, draw the picture so it's easier for you, not saying the answer is wrong tho
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– Kradec na kysmet
14 hours ago
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There are atleast 9 precious and @Weather Vane - I checked your first and it looks wrong - or do you mean 57% for precious and 43% not precious in 1 of the cases ?
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– Kradec na kysmet
13 hours ago
|
show 11 more comments
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is:
4, 5, 6, 7 must be true, while 1, 2, 3 are wrong.
Let us denote by $a, b, c$ the amount of boxes of each kind as follows:
PP PN NNa b c
First we know that $a+b+c=8$.
Now we use the $50%$ chance given by the host:
We apply Bayes' formula, where the event PP is "our box contains two precious pebbles" and P stands for "we pick a pebble from our box and it turns out to be precious".
We have $$0.5=Bbb P(P!P|P)=frac {Bbb P(P!P)}{Bbb P(P)}$$ $$0.5=Bbb P(P!P|P)=frac {a/8}{a/8+b/8cdot1/2}$$
So finally:
$$b=2a$$
Now the host also said that there are at least as many P's as N's. Therefore:
$$2a+bgeq 2c+b$$
which, given $a+b+c=8$ and $b=2a$, translates to
$$ageq 2$$
Partial conclusion
There is one configuration meeting these conditions, namely
PP PN NN2 4 2
and there is no other possibility, because $a=3$ would imply $b=6$ and $a+b$ would already be more than $8$.
We see that $4, 5, 6, 7$ are all true.
Moving on to questions 1, 2, 3:
If we have a PN box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 37cdot frac 12}_{PN}+underbrace{frac 27}_{PP}=0.5$$
If we have a PP box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 47cdot frac 12}_{PN}+underbrace{frac 17}_{PP}=frac 37$$
In the first case, the probability stays the same, while in the second it decreases. Therefore 1, 2, 3 are all wrong: it depends.
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
$endgroup$
$begingroup$
Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host
$endgroup$
– Kradec na kysmet
17 hours ago
$begingroup$
@Kradecnakysmet is that an admission that statement 8 is false?
$endgroup$
– Weather Vane
17 hours ago
1
$begingroup$
You have a contradiction in your answer. You've calculated the probability of having a PP box as 2/3. If that were truly the case, then the chance that the second pebble in that box is precious is also 2/3. There's no other way around it. So either the host is lying, the calculated probability is wrong, or your original assumption of an equal number of PP and PN boxes is incorrect.
$endgroup$
– Amorydai
13 hours ago
1
$begingroup$
Also, since I deleted my answer, here's our chat discussion, Arnaud.
$endgroup$
– Brandon_J
8 hours ago
1
$begingroup$
Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes".
$endgroup$
– ppgdev
7 hours ago
|
show 17 more comments
$begingroup$
The answer is:
4, 5, 6, 7 must be true, while 1, 2, 3 are wrong.
Let us denote by $a, b, c$ the amount of boxes of each kind as follows:
PP PN NNa b c
First we know that $a+b+c=8$.
Now we use the $50%$ chance given by the host:
We apply Bayes' formula, where the event PP is "our box contains two precious pebbles" and P stands for "we pick a pebble from our box and it turns out to be precious".
We have $$0.5=Bbb P(P!P|P)=frac {Bbb P(P!P)}{Bbb P(P)}$$ $$0.5=Bbb P(P!P|P)=frac {a/8}{a/8+b/8cdot1/2}$$
So finally:
$$b=2a$$
Now the host also said that there are at least as many P's as N's. Therefore:
$$2a+bgeq 2c+b$$
which, given $a+b+c=8$ and $b=2a$, translates to
$$ageq 2$$
Partial conclusion
There is one configuration meeting these conditions, namely
PP PN NN2 4 2
and there is no other possibility, because $a=3$ would imply $b=6$ and $a+b$ would already be more than $8$.
We see that $4, 5, 6, 7$ are all true.
Moving on to questions 1, 2, 3:
If we have a PN box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 37cdot frac 12}_{PN}+underbrace{frac 27}_{PP}=0.5$$
If we have a PP box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 47cdot frac 12}_{PN}+underbrace{frac 17}_{PP}=frac 37$$
In the first case, the probability stays the same, while in the second it decreases. Therefore 1, 2, 3 are all wrong: it depends.
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
$endgroup$
$begingroup$
Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host
$endgroup$
– Kradec na kysmet
17 hours ago
$begingroup$
@Kradecnakysmet is that an admission that statement 8 is false?
$endgroup$
– Weather Vane
17 hours ago
1
$begingroup$
You have a contradiction in your answer. You've calculated the probability of having a PP box as 2/3. If that were truly the case, then the chance that the second pebble in that box is precious is also 2/3. There's no other way around it. So either the host is lying, the calculated probability is wrong, or your original assumption of an equal number of PP and PN boxes is incorrect.
$endgroup$
– Amorydai
13 hours ago
1
$begingroup$
Also, since I deleted my answer, here's our chat discussion, Arnaud.
$endgroup$
– Brandon_J
8 hours ago
1
$begingroup$
Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes".
$endgroup$
– ppgdev
7 hours ago
|
show 17 more comments
$begingroup$
The answer is:
4, 5, 6, 7 must be true, while 1, 2, 3 are wrong.
Let us denote by $a, b, c$ the amount of boxes of each kind as follows:
PP PN NNa b c
First we know that $a+b+c=8$.
Now we use the $50%$ chance given by the host:
We apply Bayes' formula, where the event PP is "our box contains two precious pebbles" and P stands for "we pick a pebble from our box and it turns out to be precious".
We have $$0.5=Bbb P(P!P|P)=frac {Bbb P(P!P)}{Bbb P(P)}$$ $$0.5=Bbb P(P!P|P)=frac {a/8}{a/8+b/8cdot1/2}$$
So finally:
$$b=2a$$
Now the host also said that there are at least as many P's as N's. Therefore:
$$2a+bgeq 2c+b$$
which, given $a+b+c=8$ and $b=2a$, translates to
$$ageq 2$$
Partial conclusion
There is one configuration meeting these conditions, namely
PP PN NN2 4 2
and there is no other possibility, because $a=3$ would imply $b=6$ and $a+b$ would already be more than $8$.
We see that $4, 5, 6, 7$ are all true.
Moving on to questions 1, 2, 3:
If we have a PN box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 37cdot frac 12}_{PN}+underbrace{frac 27}_{PP}=0.5$$
If we have a PP box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 47cdot frac 12}_{PN}+underbrace{frac 17}_{PP}=frac 37$$
In the first case, the probability stays the same, while in the second it decreases. Therefore 1, 2, 3 are all wrong: it depends.
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
$endgroup$
The answer is:
4, 5, 6, 7 must be true, while 1, 2, 3 are wrong.
Let us denote by $a, b, c$ the amount of boxes of each kind as follows:
PP PN NNa b c
First we know that $a+b+c=8$.
Now we use the $50%$ chance given by the host:
We apply Bayes' formula, where the event PP is "our box contains two precious pebbles" and P stands for "we pick a pebble from our box and it turns out to be precious".
We have $$0.5=Bbb P(P!P|P)=frac {Bbb P(P!P)}{Bbb P(P)}$$ $$0.5=Bbb P(P!P|P)=frac {a/8}{a/8+b/8cdot1/2}$$
So finally:
$$b=2a$$
Now the host also said that there are at least as many P's as N's. Therefore:
$$2a+bgeq 2c+b$$
which, given $a+b+c=8$ and $b=2a$, translates to
$$ageq 2$$
Partial conclusion
There is one configuration meeting these conditions, namely
PP PN NN2 4 2
and there is no other possibility, because $a=3$ would imply $b=6$ and $a+b$ would already be more than $8$.
We see that $4, 5, 6, 7$ are all true.
Moving on to questions 1, 2, 3:
If we have a PN box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 37cdot frac 12}_{PN}+underbrace{frac 27}_{PP}=0.5$$
If we have a PP box and we swap, then the probability that the next pebble we pick is a P is $$underbrace{frac 47cdot frac 12}_{PN}+underbrace{frac 17}_{PP}=frac 37$$
In the first case, the probability stays the same, while in the second it decreases. Therefore 1, 2, 3 are all wrong: it depends.
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
edited 8 hours ago
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
answered 17 hours ago
Arnaud MortierArnaud Mortier
1,285420
1,285420
$begingroup$
Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host
$endgroup$
– Kradec na kysmet
17 hours ago
$begingroup$
@Kradecnakysmet is that an admission that statement 8 is false?
$endgroup$
– Weather Vane
17 hours ago
1
$begingroup$
You have a contradiction in your answer. You've calculated the probability of having a PP box as 2/3. If that were truly the case, then the chance that the second pebble in that box is precious is also 2/3. There's no other way around it. So either the host is lying, the calculated probability is wrong, or your original assumption of an equal number of PP and PN boxes is incorrect.
$endgroup$
– Amorydai
13 hours ago
1
$begingroup$
Also, since I deleted my answer, here's our chat discussion, Arnaud.
$endgroup$
– Brandon_J
8 hours ago
1
$begingroup$
Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes".
$endgroup$
– ppgdev
7 hours ago
|
show 17 more comments
$begingroup$
Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host
$endgroup$
– Kradec na kysmet
17 hours ago
$begingroup$
@Kradecnakysmet is that an admission that statement 8 is false?
$endgroup$
– Weather Vane
17 hours ago
1
$begingroup$
You have a contradiction in your answer. You've calculated the probability of having a PP box as 2/3. If that were truly the case, then the chance that the second pebble in that box is precious is also 2/3. There's no other way around it. So either the host is lying, the calculated probability is wrong, or your original assumption of an equal number of PP and PN boxes is incorrect.
$endgroup$
– Amorydai
13 hours ago
1
$begingroup$
Also, since I deleted my answer, here's our chat discussion, Arnaud.
$endgroup$
– Brandon_J
8 hours ago
1
$begingroup$
Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes".
$endgroup$
– ppgdev
7 hours ago
$begingroup$
Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host
$endgroup$
– Kradec na kysmet
17 hours ago
$begingroup$
Precious pebbles are >= Not precious; Chance for 2nd precious pebble in our box is 50%, as stated by the host
$endgroup$
– Kradec na kysmet
17 hours ago
$begingroup$
@Kradecnakysmet is that an admission that statement 8 is false?
$endgroup$
– Weather Vane
17 hours ago
$begingroup$
@Kradecnakysmet is that an admission that statement 8 is false?
$endgroup$
– Weather Vane
17 hours ago
1
1
$begingroup$
You have a contradiction in your answer. You've calculated the probability of having a PP box as 2/3. If that were truly the case, then the chance that the second pebble in that box is precious is also 2/3. There's no other way around it. So either the host is lying, the calculated probability is wrong, or your original assumption of an equal number of PP and PN boxes is incorrect.
$endgroup$
– Amorydai
13 hours ago
$begingroup$
You have a contradiction in your answer. You've calculated the probability of having a PP box as 2/3. If that were truly the case, then the chance that the second pebble in that box is precious is also 2/3. There's no other way around it. So either the host is lying, the calculated probability is wrong, or your original assumption of an equal number of PP and PN boxes is incorrect.
$endgroup$
– Amorydai
13 hours ago
1
1
$begingroup$
Also, since I deleted my answer, here's our chat discussion, Arnaud.
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
Also, since I deleted my answer, here's our chat discussion, Arnaud.
$endgroup$
– Brandon_J
8 hours ago
1
1
$begingroup$
Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes".
$endgroup$
– ppgdev
7 hours ago
$begingroup$
Arnaud Mortier, I agree with @JaapScherphuis. Statement 3 is correct. It is not about "what will happen" which "depends". It is about comparing probabilities of a specific event in two scenarios. And, if you follow your own calculations, you should see that if we swap the boxes the probability of picking up a precious pebble diminishes".
$endgroup$
– ppgdev
7 hours ago
|
show 17 more comments
$begingroup$
The selection procedure ensures that each of the 16 stones has an equal chance of being picked by the host.
Once his pick is revealed to be precious, it is still equally likely to be any one of the precious stones in the boxes. For the host's statement to be true, there must be the same number of precious stones with a precious stone partner as there are with a non-precious partner.
This means that for every box with two precious stones (which obviously both have a precious partner) there must be two boxes with a precious stone partnered with a non-precious stone.
This leaves two possibilities:
1xPP, 2xPN, 5xNN
2xPP, 4xPN, 2xNN
We are given that the number of precious stones is at least as many as the non-precious ones, so we must be in the second case:
2xPP, 4xPN, 2xNN
The other boxes have 6, or 7 precious stones in them, depending on whether the current box is PP or PN. You do not know which of these is the case, though you do know they are equally probable because the host said so. There are therefore on average 6.5 precious stones in the other boxes, and when switching boxes you have a probability of 6.5/14, or less than 50% to pick a precious stone next. Therefore statements 1 and 2 are false, 3 is true.
Statements 4-7 are easily seen to be true, and 8 and 9 are false.
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
$endgroup$
2
$begingroup$
2xPP + 4xPN doesn't make it 50% chance
$endgroup$
– Kradec na kysmet
16 hours ago
1
$begingroup$
@Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance.
$endgroup$
– Amorydai
13 hours ago
$begingroup$
@Kradecnakysmet yes it does. See my answer for another way to look at it.
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
@Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer.
$endgroup$
– Arnaud Mortier
10 hours ago
add a comment |
$begingroup$
The selection procedure ensures that each of the 16 stones has an equal chance of being picked by the host.
Once his pick is revealed to be precious, it is still equally likely to be any one of the precious stones in the boxes. For the host's statement to be true, there must be the same number of precious stones with a precious stone partner as there are with a non-precious partner.
This means that for every box with two precious stones (which obviously both have a precious partner) there must be two boxes with a precious stone partnered with a non-precious stone.
This leaves two possibilities:
1xPP, 2xPN, 5xNN
2xPP, 4xPN, 2xNN
We are given that the number of precious stones is at least as many as the non-precious ones, so we must be in the second case:
2xPP, 4xPN, 2xNN
The other boxes have 6, or 7 precious stones in them, depending on whether the current box is PP or PN. You do not know which of these is the case, though you do know they are equally probable because the host said so. There are therefore on average 6.5 precious stones in the other boxes, and when switching boxes you have a probability of 6.5/14, or less than 50% to pick a precious stone next. Therefore statements 1 and 2 are false, 3 is true.
Statements 4-7 are easily seen to be true, and 8 and 9 are false.
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
$endgroup$
2
$begingroup$
2xPP + 4xPN doesn't make it 50% chance
$endgroup$
– Kradec na kysmet
16 hours ago
1
$begingroup$
@Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance.
$endgroup$
– Amorydai
13 hours ago
$begingroup$
@Kradecnakysmet yes it does. See my answer for another way to look at it.
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
@Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer.
$endgroup$
– Arnaud Mortier
10 hours ago
add a comment |
$begingroup$
The selection procedure ensures that each of the 16 stones has an equal chance of being picked by the host.
Once his pick is revealed to be precious, it is still equally likely to be any one of the precious stones in the boxes. For the host's statement to be true, there must be the same number of precious stones with a precious stone partner as there are with a non-precious partner.
This means that for every box with two precious stones (which obviously both have a precious partner) there must be two boxes with a precious stone partnered with a non-precious stone.
This leaves two possibilities:
1xPP, 2xPN, 5xNN
2xPP, 4xPN, 2xNN
We are given that the number of precious stones is at least as many as the non-precious ones, so we must be in the second case:
2xPP, 4xPN, 2xNN
The other boxes have 6, or 7 precious stones in them, depending on whether the current box is PP or PN. You do not know which of these is the case, though you do know they are equally probable because the host said so. There are therefore on average 6.5 precious stones in the other boxes, and when switching boxes you have a probability of 6.5/14, or less than 50% to pick a precious stone next. Therefore statements 1 and 2 are false, 3 is true.
Statements 4-7 are easily seen to be true, and 8 and 9 are false.
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
$endgroup$
The selection procedure ensures that each of the 16 stones has an equal chance of being picked by the host.
Once his pick is revealed to be precious, it is still equally likely to be any one of the precious stones in the boxes. For the host's statement to be true, there must be the same number of precious stones with a precious stone partner as there are with a non-precious partner.
This means that for every box with two precious stones (which obviously both have a precious partner) there must be two boxes with a precious stone partnered with a non-precious stone.
This leaves two possibilities:
1xPP, 2xPN, 5xNN
2xPP, 4xPN, 2xNN
We are given that the number of precious stones is at least as many as the non-precious ones, so we must be in the second case:
2xPP, 4xPN, 2xNN
The other boxes have 6, or 7 precious stones in them, depending on whether the current box is PP or PN. You do not know which of these is the case, though you do know they are equally probable because the host said so. There are therefore on average 6.5 precious stones in the other boxes, and when switching boxes you have a probability of 6.5/14, or less than 50% to pick a precious stone next. Therefore statements 1 and 2 are false, 3 is true.
Statements 4-7 are easily seen to be true, and 8 and 9 are false.
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
answered 16 hours ago
Jaap ScherphuisJaap Scherphuis
16.2k12772
16.2k12772
2
$begingroup$
2xPP + 4xPN doesn't make it 50% chance
$endgroup$
– Kradec na kysmet
16 hours ago
1
$begingroup$
@Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance.
$endgroup$
– Amorydai
13 hours ago
$begingroup$
@Kradecnakysmet yes it does. See my answer for another way to look at it.
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
@Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer.
$endgroup$
– Arnaud Mortier
10 hours ago
add a comment |
2
$begingroup$
2xPP + 4xPN doesn't make it 50% chance
$endgroup$
– Kradec na kysmet
16 hours ago
1
$begingroup$
@Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance.
$endgroup$
– Amorydai
13 hours ago
$begingroup$
@Kradecnakysmet yes it does. See my answer for another way to look at it.
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
@Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer.
$endgroup$
– Arnaud Mortier
10 hours ago
2
2
$begingroup$
2xPP + 4xPN doesn't make it 50% chance
$endgroup$
– Kradec na kysmet
16 hours ago
$begingroup$
2xPP + 4xPN doesn't make it 50% chance
$endgroup$
– Kradec na kysmet
16 hours ago
1
1
$begingroup$
@Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance.
$endgroup$
– Amorydai
13 hours ago
$begingroup$
@Kradecnakysmet Actually, the more I think about it, the more I believe that is exactly what's going on. Given that the first pebble pulled was precious, there's twice the amount of chance that the box had two precious pebbles in it. 2xPP + 4xPN has 8 ways of pulling a precious pebble out. 4 of those ways will have another precious pebble in the box, and 4 will not, giving you a 50% chance.
$endgroup$
– Amorydai
13 hours ago
$begingroup$
@Kradecnakysmet yes it does. See my answer for another way to look at it.
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
@Kradecnakysmet yes it does. See my answer for another way to look at it.
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
@Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer.
$endgroup$
– Arnaud Mortier
10 hours ago
$begingroup$
@Jaap I agree with everything except the last part. I don't think you should think in terms of average. See the last part of my answer.
$endgroup$
– Arnaud Mortier
10 hours ago
add a comment |
$begingroup$
There are two arrangements that satify the given facts:
A. 4 boxes with two precious pebbles, 4 boxes with one of each
B. 3 boxes with two precious pebbles, 3 with one of each, 2 with two non-precious
Because
There are at least 8 precious pebbles
Half of the boxes with a precious pebble contain another, half do not.
So for each statement:
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is higher.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
FALSE: in case A it is either more or less.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is lower.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
FALSE: in both solutions.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
FALSE: this is only true in one solution.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
FALSE: there are more precious pebbles in both solutions.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
FALSE: in both solutions.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
FALSE: there are two arrangements that satisfy the condition.
None of the above.
TRUE:
So my answer is
Statement 9 is the only true statement.
answered 18 hours ago
Weather VaneWeather Vane
1,55219
$endgroup$
$begingroup$
For it to be true it has to be true for all conditions
$endgroup$
– Kradec na kysmet
16 hours ago
$begingroup$
You have to make sure it's true for all conditions
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
There's more than 1 condition, check it carefully
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
Recheck your answers, I see lots of mistakes, draw the picture so it's easier for you, not saying the answer is wrong tho
$endgroup$
– Kradec na kysmet
14 hours ago
$begingroup$
There are atleast 9 precious and @Weather Vane - I checked your first and it looks wrong - or do you mean 57% for precious and 43% not precious in 1 of the cases ?
$endgroup$
– Kradec na kysmet
13 hours ago
|
show 11 more comments
$begingroup$
There are two arrangements that satify the given facts:
A. 4 boxes with two precious pebbles, 4 boxes with one of each
B. 3 boxes with two precious pebbles, 3 with one of each, 2 with two non-precious
Because
There are at least 8 precious pebbles
Half of the boxes with a precious pebble contain another, half do not.
So for each statement:
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is higher.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
FALSE: in case A it is either more or less.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is lower.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
FALSE: in both solutions.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
FALSE: this is only true in one solution.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
FALSE: there are more precious pebbles in both solutions.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
FALSE: in both solutions.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
FALSE: there are two arrangements that satisfy the condition.
None of the above.
TRUE:
So my answer is
Statement 9 is the only true statement.
answered 18 hours ago
Weather VaneWeather Vane
1,55219
$endgroup$
$begingroup$
For it to be true it has to be true for all conditions
$endgroup$
– Kradec na kysmet
16 hours ago
$begingroup$
You have to make sure it's true for all conditions
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
There's more than 1 condition, check it carefully
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
Recheck your answers, I see lots of mistakes, draw the picture so it's easier for you, not saying the answer is wrong tho
$endgroup$
– Kradec na kysmet
14 hours ago
$begingroup$
There are atleast 9 precious and @Weather Vane - I checked your first and it looks wrong - or do you mean 57% for precious and 43% not precious in 1 of the cases ?
$endgroup$
– Kradec na kysmet
13 hours ago
|
show 11 more comments
$begingroup$
There are two arrangements that satify the given facts:
A. 4 boxes with two precious pebbles, 4 boxes with one of each
B. 3 boxes with two precious pebbles, 3 with one of each, 2 with two non-precious
Because
There are at least 8 precious pebbles
Half of the boxes with a precious pebble contain another, half do not.
So for each statement:
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is higher.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
FALSE: in case A it is either more or less.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is lower.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
FALSE: in both solutions.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
FALSE: this is only true in one solution.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
FALSE: there are more precious pebbles in both solutions.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
FALSE: in both solutions.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
FALSE: there are two arrangements that satisfy the condition.
None of the above.
TRUE:
So my answer is
Statement 9 is the only true statement.
answered 18 hours ago
Weather VaneWeather Vane
1,55219
$endgroup$
There are two arrangements that satify the given facts:
A. 4 boxes with two precious pebbles, 4 boxes with one of each
B. 3 boxes with two precious pebbles, 3 with one of each, 2 with two non-precious
Because
There are at least 8 precious pebbles
Half of the boxes with a precious pebble contain another, half do not.
So for each statement:
If we swapped the boxes, we would have higher chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is higher.
If we swapped the boxes, we would have same chance of picking up precious pebble next.
FALSE: in case A it is either more or less.
If we swapped the boxes, we would have lower chance of picking up precious pebble next.
FALSE: in case A, there was 1/2 (0.500) chance of picking a precious pebble. If you take out that box there will be either a 3/7 (0.428) or a 4/7 (0.571) chance. Only one of those is lower.
In the beginning of the game, there've been exactly 2 boxes with 2 precious pebbles each.
FALSE: in both solutions.
In the beginning of the game, half the boxes have had 1 precious and 1 not precious pebble.
FALSE: this is only true in one solution.
In the beginning of the game, the precious and not precious pebbles have been equal in count.
FALSE: there are more precious pebbles in both solutions.
In the beginning of the game, the amount of boxes with 2 precious pebbles is equal to the amount of boxes with 2 not precious pebbles.
FALSE: in both solutions.
The TV host has gone wrong. There's no way that there's 50% chance for 2nd precious pebble in our box, as there is unequal amount of pebbles left in the game.
FALSE: there are two arrangements that satisfy the condition.
None of the above.
TRUE:
So my answer is
Statement 9 is the only true statement.
answered 18 hours ago
Weather VaneWeather Vane
1,55219
edited 13 hours ago
answered 18 hours ago
Weather VaneWeather Vane
1,55219
answered 18 hours ago
Weather VaneWeather Vane
1,55219
answered 18 hours ago
Weather VaneWeather Vane
1,55219
1,55219
$begingroup$
For it to be true it has to be true for all conditions
$endgroup$
– Kradec na kysmet
16 hours ago
$begingroup$
You have to make sure it's true for all conditions
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
There's more than 1 condition, check it carefully
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
Recheck your answers, I see lots of mistakes, draw the picture so it's easier for you, not saying the answer is wrong tho
$endgroup$
– Kradec na kysmet
14 hours ago
$begingroup$
There are atleast 9 precious and @Weather Vane - I checked your first and it looks wrong - or do you mean 57% for precious and 43% not precious in 1 of the cases ?
$endgroup$
– Kradec na kysmet
13 hours ago
|
show 11 more comments
$begingroup$
For it to be true it has to be true for all conditions
$endgroup$
– Kradec na kysmet
16 hours ago
$begingroup$
You have to make sure it's true for all conditions
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
There's more than 1 condition, check it carefully
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
Recheck your answers, I see lots of mistakes, draw the picture so it's easier for you, not saying the answer is wrong tho
$endgroup$
– Kradec na kysmet
14 hours ago
$begingroup$
There are atleast 9 precious and @Weather Vane - I checked your first and it looks wrong - or do you mean 57% for precious and 43% not precious in 1 of the cases ?
$endgroup$
– Kradec na kysmet
13 hours ago
$begingroup$
For it to be true it has to be true for all conditions
$endgroup$
– Kradec na kysmet
16 hours ago
$begingroup$
For it to be true it has to be true for all conditions
$endgroup$
– Kradec na kysmet
16 hours ago
$begingroup$
You have to make sure it's true for all conditions
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
You have to make sure it's true for all conditions
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
There's more than 1 condition, check it carefully
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
There's more than 1 condition, check it carefully
$endgroup$
– Kradec na kysmet
15 hours ago
$begingroup$
Recheck your answers, I see lots of mistakes, draw the picture so it's easier for you, not saying the answer is wrong tho
$endgroup$
– Kradec na kysmet
14 hours ago
$begingroup$
Recheck your answers, I see lots of mistakes, draw the picture so it's easier for you, not saying the answer is wrong tho
$endgroup$
– Kradec na kysmet
14 hours ago
$begingroup$
There are atleast 9 precious and @Weather Vane - I checked your first and it looks wrong - or do you mean 57% for precious and 43% not precious in 1 of the cases ?
$endgroup$
– Kradec na kysmet
13 hours ago
$begingroup$
There are atleast 9 precious and @Weather Vane - I checked your first and it looks wrong - or do you mean 57% for precious and 43% not precious in 1 of the cases ?
$endgroup$
– Kradec na kysmet
13 hours ago
|
show 11 more comments
Kradec na kysmet is a new contributor. Be nice, and check out our Code of Conduct.
Kradec na kysmet is a new contributor. Be nice, and check out our Code of Conduct.
Kradec na kysmet is a new contributor. Be nice, and check out our Code of Conduct.
Kradec na kysmet is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
This is somewhat related to the somewhat annoying but extremely fascinating Monty Hall problem.
$endgroup$
– Brandon_J
7 hours ago