# Probability problem

## Homework Statement

Two fair dices are thrown and the product of the numbers on the dice is recorded, given that one die lands on 2, find the probability that the product on the dice is exactly 6

## Homework Equations

P(A|B)=P(AnB)/P(B)

## The Attempt at a Solution

I drew the table of the products The intersection is 2/36
But how can I find the probability of landing on 2?

Related Precalculus Mathematics Homework Help News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
But how can I find the probability of landing on 2?
Looking at your outcome table: How many combination of dice rolls end up with at least one 2?

Looking at your outcome table: How many combination of dice rolls end up with at least one 2?
2 combinations? 2 and 1, and 1 and 2

Orodruin
Staff Emeritus
Homework Helper
Gold Member
What about 2 and 3 or 2 and 4? Those have at least one 2.

Edit: Let me put it this way. How many combinations have a 2 on Die 1? How many combinations have a 2 on Die 2?

What about 2 and 3 or 2 and 4? Those have at least one 2.

Edit: Let me put it this way. How many combinations have a 2 on Die 1? How many combinations have a 2 on Die 2?
So total of 10 combinations right?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
So total of 10 combinations right?
This depends on how you interpret the problem formulation (which is not clear). If the formulation is that only one die should show a 2, then 10 is correct. If the formulation also allows both dice showing 2, then you get an additional possible outcome.

This depends on how you interpret the problem formulation (which is not clear). If the formulation is that only one die should show a 2, then 10 is correct. If the formulation also allows both dice showing 2, then you get an additional possible outcome.
How do I know if I count for 2 dices or just one? Or in other way why I can't count for only 1 die landing on 2

Orodruin
Staff Emeritus
Homework Helper
Gold Member
How do I know if I count for 2 dices or just one? Or in other way why I can't count for only 1 die landing on 2
This would depend on problem formulation. As you have formulated the problem above, I would interpret it as at least one die landing on 2. Other possibilities would include:
• Dice 1 landing on 2 and Dice 2 arbitrary.
• Only one die landing on 2 and the other on a different number.
I would say it is up to the problem maker to formulate the problem such that it is unambiguous what is intended. For the case of at least one die landing on 2, the possibilities are the 10 possibilities I suspect you counted (2x and x2 where x is not equal to 2) plus the possibility of both dice landing on 2.

This would depend on problem formulation. As you have formulated the problem above, I would interpret it as at least one die landing on 2. Other possibilities would include:
• Dice 1 landing on 2 and Dice 2 arbitrary.
• Only one die landing on 2 and the other on a different number.
I would say it is up to the problem maker to formulate the problem such that it is unambiguous what is intended. For the case of at least one die landing on 2, the possibilities are the 10 possibilities I suspect you counted (2x and x2 where x is not equal to 2) plus the possibility of both dice landing on 2.
That explains alot thanks
The problem is there is a similar question where it says that there're two spinners(1 to 4), given that at least one spinner lands on 3, and they counted the outcome of one spinner only! :\

haruspex
Homework Helper
Gold Member
I would say it is up to the problem maker to formulate the problem such that it is unambiguous what is intended. For the case of at least one die landing on 2, the possibilities are the 10 possibilities I suspect you counted (2x and x2 where x is not equal to 2) plus the possibility of both dice landing on 2.
I have a slightly different view. It says
given that one die lands on 2
I agree that, almost surely, the poser intended to convey
given that at least one die lands on 2
but that is not what it says. There are two other interpretations of the actual words that are at least as reasonable:
- that exactly one die landed on 2, producing 10 cases
- that a specific die landed on 2... maybe the red one, or the first rolled, or the first the poser looked at... producing only 6 cases.

haruspex
Homework Helper
Gold Member
The problem is there is a similar question where it says that there're two spinners(1 to 4), given that at least one spinner lands on 3, and they counted the outcome of one spinner only! :\
I couldn't understand that description. Would you post that question exactly as worded please?

vela
Staff Emeritus
Homework Helper

## Homework Statement

Two fair dices are thrown and the product of the numbers on the dice is recorded.
How do I know if I count for 2 dices or just one? Or in other way why I can't count for only 1 die landing on 2
Just thought you might want to know that dice is the plural of die, so it's "two dice" not "two dices". :)

Orodruin
Staff Emeritus
Homework Helper
Gold Member
but that is not what it says. There are two other interpretations of the actual words that are at least as reasonable:
- that exactly one die landed on 2, producing 10 cases
- that a specific die landed on 2... maybe the red one, or the first rolled, or the first the poser looked at... producing only 6 cases.
But those two cases are exactly the cases included in my post #8 as other possibilities. This is why I have been pointing out that the formulation is ambiguous.

haruspex
Homework Helper
Gold Member
But those two cases are exactly the cases included in my post #8 as other possibilities. This is why I have been pointing out that the formulation is ambiguous.
I'm very sorry - for some reason I thought you'd only only covered the exactly one alternative. I must have read it too quickly.

Homework Helper
" given that one die lands on 2, find the probability that the product on the dice is exactly 6"

would typically be taken to mean "at least one die" - if a single die were intended then it would read "exactly one"
For the OP: at least one means or or the other or possibly both: would it be possible for the sum to be 6 if both of the dice end up with 2 dots showing?

haruspex
Homework Helper
Gold Member
" given that one die lands on 2, find the probability that the product on the dice is exactly 6"

would typically be taken to mean "at least one die" - if a single die were intended then it would read "exactly one"
For the OP: at least one means or or the other or possibly both: would it be possible for the sum to be 6 if both of the dice end up with 2 dots showing?
The only reason it typically means that in probability questions is that it makes it an interesting question. In everyday usage, that is about the least likely of the three possibilities.
If a friend were to say to you "I have two children. One is a girl", you would assume the other is a boy.
Alternatively, suppose he only told you that he had two children, and happened to have a girl with him at the time, you might report back to your partner that "Albert has two children. One of them is a girl". You might even say "at least one is a girl", yet the probability that the other is a girl remains 1/2, because you only know about one child.
So while the required answer here is almost surely 2/11, it is a badly worded question. The English language is not good at conveying logical relationships, so a conversational style is fraught with traps for the unwary problem setter.

Homework Helper
"The only reason it typically means that in probability questions is that it makes it an interesting question."
That is the context - when we (I and others) write problems for tests, or text, the specific case (exactly 1) is the one that must be stated.

"So while the required answer here is almost surely 2/11, it is a badly worded question."
In everyday discussion you would be correct. In the general academic setting - no: it is the way things are. You are mixing two cultures. (You can make a very good case that academics in general, and textbook authors in particular, need to be more diligent in their writing and I would not disagree with you. That's not how the current situation is framed.)

haruspex
Homework Helper
Gold Member
"The only reason it typically means that in probability questions is that it makes it an interesting question."
That is the context
The context is that it must be an interesting question? How many students are taught that, and how would they know when to apply it?
No, of course you mean that the context is probability questions set in courses, but if that's your defence then the syllabus had better include learning this arcane usage.
when we (I and others) write problems for tests, or text, the specific case (exactly 1) is the one that must be stated.
In everyday discussion you would be correct. In the general academic setting - no: it is the way things are.
I don't accept any of that as excusing the problem setter. It is a practice that has become accepted through custom, but it creates an unnecessary hurdle for every generation of students that has nothing to do with testing their understanding of probability.

An Australian HSC question about 15 years ago set the classic question about two balls, one of which was red. It framed it as Joe having the two balls behind his back, drawn from four red and four black say, but dropped one, and that was seen to be red. Clearly this framing doesn't work. Identifying a specific ball as red means the other ball has a probability 3/7 of being red, not the 3/11 that was marked correct. This customary usage is so pernicious it even catches out exam setters.
You are mixing two cultures.
No, that's my complaint: the problem setter mixes two cultures by adopting a conversational style.
For a mathematical context there's mathematical style, something like "the set of numbers shown by the dice includes 2". A conversational style is OK, but it is up to the question setter ensure that no ambiguity arises.

Interestingly, in writing my previous post I came to realise that even saying "at least one" does not really solve it. Consider the example I gave about meeting Albert and his little girl:
you might report back to your partner that "Albert has two children. At least one of them is a girl".​
Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?

Ray Vickson
Homework Helper
Dearly Missed
The context is that it must be an interesting question? How many students are taught that, and how would they know when to apply it?
No, of course you mean that the context is probability questions set in courses, but if that's your defence then the syllabus had better include learning this arcane usage.

I don't accept any of that as excusing the problem setter. It is a practice that has become accepted through custom, but it creates an unnecessary hurdle for every generation of students that has nothing to do with testing their understanding of probability.

An Australian HSC question about 15 years ago set the classic question about two balls, one of which was red. It framed it as Joe having the two balls behind his back, drawn from four red and four black say, but dropped one, and that was seen to be red. Clearly this framing doesn't work. Identifying a specific ball as red means the other ball has a probability 3/7 of being red, not the 3/11 that was marked correct. This customary usage is so pernicious it even catches out exam setters.

No, that's my complaint: the problem setter mixes two cultures by adopting a conversational style.
For a mathematical context there's mathematical style, something like "the set of numbers shown by the dice includes 2". A conversational style is OK, but it is up to the question setter ensure that no ambiguity arises.

Interestingly, in writing my previous post I came to realise that even saying "at least one" does not really solve it. Consider the example I gave about meeting Albert and his little girl:
you might report back to your partner that "Albert has two children. At least one of them is a girl".​
Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?
The context is that it must be an interesting question? How many students are taught that, and how would they know when to apply it?
No, of course you mean that the context is probability questions set in courses, but if that's your defence then the syllabus had better include learning this arcane usage.

I don't accept any of that as excusing the problem setter. It is a practice that has become accepted through custom, but it creates an unnecessary hurdle for every generation of students that has nothing to do with testing their understanding of probability.

An Australian HSC question about 15 years ago set the classic question about two balls, one of which was red. It framed it as Joe having the two balls behind his back, drawn from four red and four black say, but dropped one, and that was seen to be red. Clearly this framing doesn't work. Identifying a specific ball as red means the other ball has a probability 3/7 of being red, not the 3/11 that was marked correct. This customary usage is so pernicious it even catches out exam setters.

No, that's my complaint: the problem setter mixes two cultures by adopting a conversational style.
For a mathematical context there's mathematical style, something like "the set of numbers shown by the dice includes 2". A conversational style is OK, but it is up to the question setter ensure that no ambiguity arises.

Interestingly, in writing my previous post I came to realise that even saying "at least one" does not really solve it. Consider the example I gave about meeting Albert and his little girl:
you might report back to your partner that "Albert has two children. At least one of them is a girl".​
Your partner cannot estimate the probability that the other is also a girl without knowing how you knew that - did you see both children or only one? Or did Albert say something that implied one was a girl, and if so what?
If you just report what you saw then one can make a good case that the probability the other child is a girl is 1/3 (because if you see a girl it eliminates the case BB, leaving three equally-likely cases GB, BG, GG--in order youngest-oldest). However, if you know, for example, that the child you saw is the oldest (or youngest) that changes the probability of the other being a girl to 1/2 (assuming the usual independence and equal birth rates, etc.)

haruspex
Homework Helper
Gold Member
If you just report what you saw then one can make a good case that the probability the other child is a girl is 1/3 (because if you see a girl it eliminates the case BB, leaving three equally-likely cases GB, BG, GG--in order youngest-oldest). However, if you know, for example, that the child you saw is the oldest (or youngest) that changes the probability of the other being a girl to 1/2 (assuming the usual independence and equal birth rates, etc.)
Well, Ray, I did not expect you to fall into that trap. You've illustrated the issue nicely.
If you see one child and that child is a girl, you have eliminated two equally likely cases:
Code:
  case; child you see; other child
1        g           g
2        g           b
3        b           g
4        b           b
Seeing a girl eliminates 3 and 4.
Any arbitrary reason for selecting a particular child of whom to state the gender brings the probability back to 1/2.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Well, Ray, I did not expect you to fall into that trap. You've illustrated the issue nicely.
If you see one child and that child is a girl, you have eliminated two equally likely cases:
Code:
  case; child you see; other child
1        g           g
2        g           b
3        b           g
4        b           b
Seeing a girl eliminates 3 and 4.
Any arbitrary reason for selecting a particular child of whom to state the gender brings the probability back to 1/2.
Or, without the table, using Bayes' theorem:
P(2girls | see girl) = P( see girl | 2 girls) P(2 girls) / P(see girl) = 1*(1/4)/(1/2) = 1/2.

Ray Vickson
Homework Helper
Dearly Missed
Well, Ray, I did not expect you to fall into that trap. You've illustrated the issue nicely.
If you see one child and that child is a girl, you have eliminated two equally likely cases:
Code:
  case; child you see; other child
1        g           g
2        g           b
3        b           g
4        b           b
Seeing a girl eliminates 3 and 4.
Any arbitrary reason for selecting a particular child of whom to state the gender brings the probability back to 1/2.
Look at it differently. The three relevant outcomes for what the children could be are BG, GB, GG (where these are in the order oldest-youngest). However, we need to look also at the "seeing" aspect, so this produces a 6-element sample space (the three where we see the youngest and the three where we see the oldest). Among these 6 sample points, two of them have two girls, giving P(2 girls) = 1/3, as I suggested before. Of course, as soon as you know you are seeing the oldest (or the youngest) that collapses the resulting sample space into two elements, giving you the 1/2 probability. And, as you say, we need to assume that the announcement of gender was random, which in reality might be difficult to guarantee. While I am not completely sure, I suspect that we could devise some types of mixed announcement strategies that produce a 2G probability between 1/3 and 1/2.

I don't think this problem is substantially different from the following coin-tossing example: "Your friend tossed a (fair) coin twice and announced that he got at least one head. What is the probability that he got two heads?" If you have no information about the 'position' of the announced head (first vs. second) the probability is 1/3, but if you are told the position it reverts to 1/2. If you have some possibly imperfect information about the position, perhaps a result between 1/3 and 1/2 can be obtained.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Look at it differently. The three relevant outcomes for what the children could be are BG, GB, GG (where these are in the order oldest-youngest). However, we need to look also at the "seeing" aspect, so this produces a 6-element sample space (the three where we see the youngest and the three where we see the oldest). Among these 6 sample points, two of them have two girls, giving P(2 girls) = 1/3, as I suggested before. Of course, as soon as you know you are seeing the oldest (or the youngest) that collapses the resulting sample space into two elements, giving you the 1/2 probability. And, as you say, we need to assume that the announcement of gender was random, which in reality might be difficult to guarantee. While I am not completely sure, I suspect that we could devise some types of mixed announcement strategies that produce a 2G probability between 1/3 and 1/2.

I don't think this problem is substantially different from the following coin-tossing example: "Your friend tossed a (fair) coin twice and announced that he got at least one head. What is the probability that he got two heads?" If you have no information about the 'position' of the announced head (first vs. second) the probability is 1/3, but if you are told the position it reverts to 1/2. If you have some possibly imperfect information about the position, perhaps a result between 1/3 and 1/2 can be obtained.
Even if you randomly chose which child to announce the gender of (without selection bias), the probability becomes 1/2. This becomes evident from the argument using Bayes theorem. In order to get 1/3, you need to introduce a selection bias towards girls so that P(see girl) = 3/4. One possibility being that your friend was told to bring a girl if he has one.

Ray Vickson
Homework Helper
Dearly Missed
Even if you randomly chose which child to announce the gender of (without selection bias), the probability becomes 1/2. This becomes evident from the argument using Bayes theorem. In order to get 1/3, you need to introduce a selection bias towards girls so that P(see girl) = 3/4. One possibility being that your friend was told to bring a girl if he has one.
No selection bias was used in getting the 1/3---in fact, just the opposite: the selection was random, as I tried to explain rather carefully. Again, the Bayes results need a sample space to start with, and one must be careful not to force the result just by the way the data is presented. If my two out of six argument is wrong, I would like to know how and why.

Orodruin
Staff Emeritus