# Taking socks out of drawers, conditional probability

• I
• CGandC

#### CGandC

TL;DR Summary
x
Problem: In a dresser there are 3 drawers. In one drawer there are two black socks and one white sock, in the second drawer there are two white socks, and in the third drawer there is a black and white sock. Suppose I chose a drawer randomly ( meaning, in a uniform distribution ) and I took a sock out of it randomly. What are the odds that the sock I took out is white?

Attempt:
Denote as ## D_i ## the event in which we chose drawer ## i ##. Notice that for every drawer we pick, we also pick a sock. Hence the sample space is ## \Omega = \{ (d,s) : d \in \{ 1 , 2 , 3 \} , s \in \{ \text{ White Sock , Black sock}\} \} ##.
Note that ## \Omega = 3 \cdot ( \frac{3!}{2! \cdot 1! } + \frac{2!}{2!} + \frac{2!}{1!1!} ) = 3\cdot ( 3+1+2 ) = 18 ##.

Let ## D_i ## denote the event of choosing the ## i ##-th drawer, note that ## D_1 = \{ (1,\text{ white sock }), (1,\text{ black sock } \})## , ## D_3 = \{ (3,\text{ white sock } ),(3,\text{ black sock }) \}##, ## D_2 = \{ (2,\text{ white sock }) \}##.

Denote as ## W ## the event of choosing a white sock. Note that
## W = \{ (1,\text{ white sock }) , (2,\text{ white sock }) , (3,\text{ white sock }) \} ##
Denote as ## S ## the event of choosing a sock. Note that ## S = \Omega ##.
We'll Calculate ## P(W| D \cap S ) = P(W| D ) ##,

Note that ## D_1 \cup D_2 \cup D_3 = \Omega ##, hence
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{18} = \frac{1}{9} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{18} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{18} = \frac{1}{9} ##

## P(W|D_1) = \frac{ | W \cap D_1 |}{ |D_1|} = \frac{1}{2} ##
## P(W|D_2) = \frac{ | W \cap D_2 |}{ |D_2|} = \frac{1}{1} ##
## P(W|D_3) = \frac{ | W \cap D_3 |}{ |D_3|} = \frac{1}{2} ##

And from the law of total probability,
Thus, ## P(W| D ) = P(W|D_1)P(D_1) + P(W|D_2)P(D_2) + P(W|D_3)P(D_3) = \frac{1}{6} ##

Solution from another student which I think is the correct one judging by other answers on the web:
W - a white sock was taken out.
We'll want to calculate ## P(W) ##.
## E_i ## - we choose the i-th drawer. ( since we are talking about uniform distribution space, ## P(E_i) = 1/3 ## for ## i \in \{ 1,2,3 \} ## )
From the law of total probability, notice that
## P(W,E_1) = 1/3 ## , ## P(W,E_2) = 1 ##, ## P(W,E_3) = 1/2 ##
Hence:
## P(W) = \sum_{i=1}^3 P(W|E_i)P(E_i) = P(W|E_1)P(E_1) + P(W|E_2)P(E_2) + P(W|E_3)P(E_3) = (1/3)\cdot (1/3) + 1 \cdot (1/3) + (1/2)\cdot (1/3) = 11/18 ##

My question:

Where did my attempt fail ( the final calculation is incorrect ) ? Is it because of how I defined the sample space and the ## D_i ## ? I can't figure out, can you please help me?

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Notice that the sum of the ##D_i##s is not 1. That should tell you that something is wrong there.
If the choice of the drawer is uniform from three drawers, why aren't those probabilities 1/3?

You're right, but I still don't understand about the relation between the drawer choice and the socks in the sample set. If the probability of ## D_i ## were ## 1/3 ## each then I have don't see how the sock choosing is manifested in this problem since we don't refer to them in the sample set, rather we consider the sample set as ## \Omega = \{ 1,2,3 \} ## where the numbers are the different drawers .
How can one model the sample set as ## \Omega = \{ (d,s) : d \in \{ 1 , 2 , 3 \} , s \in \{ \text{ White Sock , Black sock}\} \}
##, and yet maintain the probability of ## D_i ## to be ## 1/3 ## ?
Because we're not talking only about choosing drawers but also about choosing socks.

You are only given the probability of a white sock given you have chosen a particular drawer. Look at how that is directly used in the other student's solution. I do not understand your calculation of the ##P(D_i)##s. Those probabilities are wrong.

• hutchphd
I just don't understand what sample space to look at, on the one hand when calculating ## P(D_i) ## then ## D_i ## is a subset of some sample space, which I think, according to the offered probabilties in the student's answer, is ## \{ 1,2,3 \} ##.
On the otherhand, when calculating ## P(D_i \cap W ) ##, we are looking at the event ## D_i \cap W ## and clearly it isn't ## \{ i\} ## , but rather something like ## \{ (i, \text{ "White Sock" }) \} ##.
So how exactly the elements of the sample space/set are supposed to look like?

On the otherhand, when calculating ## P(D_i \cap W ) ##, we are looking at the event ## D_i \cap W ## and clearly it isn't ## \{ i\} ## , but rather something like ## \{ (i, \text{ "White Sock" }) \} ##.
So how exactly the elements of the sample space/set are supposed to look like?
When you want to find ##P(D_i \cap W)## and it is not directly given, you can use either:
##P(D_i \cap W) = P(W | D_i) P(D_i)##
or
##P(D_i \cap W) = P(D_i | W) P(W)##
Chose wisely. One of these is very easy and the numbers are virtually given. The other is hard.

I can reach the answer by calculating the probabilities, but I want to understand the "behind the scenes" and that stems from me not understanding what ## \Omega ## ( the sample space ) looks like in this problem; what do its elements look like? are they ordered pairs?

There are no hard and fast rules about sample spaces, but I'd say that the sample space is the socks. Each sample is one sock. The problem is to find the probability for each sock.

Sure, you can use ordered pairs of (drawer, sock) to identify the socks. But that is just a naming convention.

Looking again at the sample space with freshened mindset, I think the natural sample space is ## \Omega = \{ (Drawer_1,White),(Drawer_1,Black), (Drawer_2,White) , (Drawer_3,White) , (Drawer_3,Black) \} ##, from here we see that
## | \Omega | = 5##, ## |D_1| =2 , |D_2| =1, |D_3| =2 ##,
Now, I indeed get that
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
And indeed ## P(D_1) +P(D_2)+P(D_3) = 1 ##.

Seems like I've calculated the sample set size wrongly in my first attempt ( and it wasen't entirely the sample set in the question since there were more elements, but that is ok since I could've attributed zero probability to those elements in our discrete probability space ).
Initially I tried to combinatorically calculate the sample space's size with the following mindset: First choose a drawer ( there are 3 options for it ) and then for the drawer, pick up a sock ( for ## D_1 ## there are 3 options, for ## D_2 ## there is 1 option, for ## D_3 ## there are 2 options, ) hence I've got ## |\Omega | = 3 \cdot ( \frac{3!}{2! \cdot 1! } + \frac{2!}{2!} + \frac{2!}{1!1!} ) = 3\cdot ( 3+1+2 ) = 18## ; but clearly this was wrong, so:

1. If ## \sum{P(E_i)} = 1 ## where ## E_i ## are some events, then does this imply that ## E_1 \cup \cdots \cup E_k = \Omega ## ?

2. How can one combinatorically correctly calculate the size of the sample set as I've tried to do ( but failed as I've got wrong result )?

Looking again at the sample space with freshened mindset, I think the natural sample space is ## \Omega = \{ (Drawer_1,White),(Drawer_1,Black), (Drawer_2,White) , (Drawer_3,White) , (Drawer_3,Black) \} ##, from here we see that
## | \Omega | = 5##, ## |D_1| =2 , |D_2| =1, |D_3| =2 ##,
Now, I indeed get that
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
And indeed ## P(D_1) +P(D_2)+P(D_3) = 1 ##.
Please define the event, ##D_1##. If it is the event that the selected sock was from drawer one, then ##P(D_1)=1/3##. The problem states clearly that the drawer is selected from three drawers with equal probability:
I chose a drawer randomly ( meaning, in a uniform distribution )
You are making this part more complicated than it should be.

• pbuk
• CGandC, sysprog and FactChecker
I would quibble that this strongly implies a sample space of socks.
Yes, in the same way as ## x = 2 + 2 ## implies a function ## f(a, b) \to c ## over a commutative group but until you can reliably solve ## x = 4 ## you are not ready for group theory.

• PeroK
So the worked solution which should gain full marks at any exam in which this is an appropriate level question without any mention of sample spaces or conditional probability is:

Probability of selecting a white sock from:
- the first drawer: ## \frac 1 3 \times \frac 1 3 = \frac 1 9 ##
- the second drawer: ## \frac 1 3 \times 1 = \frac 1 3 ##
- the third drawer: ## \frac 1 3 \times \frac 1 2 = \frac 1 6 ##

Total probability of selecting a white sock: ## \frac 1 9 + \frac 1 3 + \frac 1 6 = \frac 2 {18} + \frac 6 {18} + \frac 3 {18} = \frac {11} {18} ##.

• DrClaude and FactChecker
Thanks, I understand all of these but my question is about something else. I still don't understand why my combinatorial calculation of the sample space is wrong ( I know it's wrong but I don't understand why ). I explained in my comment above as follows
Initially I tried to combinatorically calculate the sample space's size with the following mindset: First choose a drawer ( there are 3 options for it ) and then for the drawer, pick up a sock ( for ## D_1 ## there are 3 options, for ## D_2 ## there is 1 option, for ## D_3 ## there are 2 options, ) hence I've got ## |\Omega | = 3 \cdot ( \frac{3!}{2! \cdot 1! } + \frac{2!}{2!} + \frac{2!}{1!1!} ) = 3\cdot ( 3+1+2 ) = 18## ; but clearly this was wrong, so:

How does one correctly calculate the sample space's size using combinatorial argumentation as I have tried? ( Again, I'm not talking about calculating a probability but rather calculating size of a finite set )
( which is ##
\Omega = \{ (Drawer_1,White),(Drawer_1,Black), (Drawer_2,White) , (Drawer_3,White) , (Drawer_3,Black) \}
## )

I think a calculation using trees as offered above will be a good idea, I'll do it that way and try to solve the problem again.

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How does one correctly calculate the sample space's size using combinatorial argumentation as I have tried?
By constructing a tree: the size of the sample space is the number of leaf nodes.

So the worked solution which should gain full marks at any exam in which this is an appropriate level question without any mention of sample spaces or conditional probability is:

Probability of selecting a white sock from:
- the first drawer: ## \frac 1 3 \times \frac 1 3 = \frac 1 9 ##
- the second drawer: ## \frac 1 3 \times 1 = \frac 1 3 ##
- the third drawer: ## \frac 1 3 \times \frac 1 2 = \frac 1 6 ##

Total probability of selecting a white sock: ## \frac 1 9 + \frac 1 3 + \frac 1 6 = \frac 2 {18} + \frac 6 {18} + \frac 3 {18} = \frac {11} {18} ##.
I don't know any level where these problems are done and conditional probabilities are not mentioned. Certainly, they are being used here in such an intuitive way that there is no need to name them. But a rose by any other name is still a rose. ;-)

Thanks, I understand all of these but my question is about something else. I still don't understand why my combinatorial calculation of the sample space is wrong ( I know it's wrong but I don't understand why ).
To focus in on the problem, your values of the ##P(D_i)## are all wrong. I can't follow how you got your numbers, so I can not help more than that.

I don't know any level where these problems are done and conditional probabilities are not mentioned.
Oh absolutely, I just meant that I wouldn't expect them in the solution to this problem because:
they are being used here in such an intuitive way that there is no need to name them

Another example at an appropriate level (which in the UK is Year 10/11) https://www.bbc.co.uk/bitesize/guides/zsrq6yc/revision/8

• FactChecker
Well, I can solve it using intuition and basic stuff, but I'm trying to solve it by first principles ( a more complicated way) for the sake of learning, not just solving.
Here's what I've tried to calculate the probability:

First, I've written the sample space ## \Omega = \{ (d_1,w),(d_1,b), (d_2,w) , (d_3,w) ,(d_3,b) \} ##. ( ## d_1,d_2,d_3 ## stand for drawers ##1,2,3 ## respectively and ##w,b## stand for White,Black )

Denote ## D_1,D_2,D_3 ## as the events of choosing drawers ##1,2,3 ## respectively. Denote ## W ## as the event of choosing a white sock.
Note that,
## D_1 = \{ (d_1,w),(d_1,b) \} ##, ## D_2 = \{ (d_2,w) \} ##, ## D_3 = \{ (d_3,w) ,(d_3,b) \} ##.
Note that
## W\cap D_1 = \{ (d_1,w) \} ##,## W\cap D_2 = \{ (d_2,w) \} ##,## W\cap D_3 = \{ (d_3,w) \} ##.

## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
( Note that their sum is 1 )

## P( W| D_1) = \frac{ P(W\cap D_1) }{ P(D_1) } = \frac{ \frac{1}{5} }{ \frac{2}{5} } = \frac{1}{2} ##
## P( W| D_2) = \frac{ P(W\cap D_2) }{ P(D_2) } = \frac{ \frac{1}{5} }{ \frac{1}{5} } = 1 ##
## P( W| D_3) = \frac{ P(W\cap D_3) }{ P(D_3) } = \frac{ \frac{1}{5} }{ \frac{2}{5} } = \frac{1}{2} ##

Hence, from law of total probability we get:
## P(W ) = P(W|D_1)P(D_1) + P(W|D_2)P(D_2) + P(W|D_3)P(D_3) = 1/2 \cdot 2/5 + 1 \cdot 1/5 + 1/2 \cdot 2/5 = 3/5 ##

I still don't see what I've done wrong in my above analysis, if you have any idea then that might pinpoint my misunderstanding.

Well, I can solve it using intuition and basic stuff, but I'm trying to solve it by first principles ( a more complicated way) for the sake of learning, not just solving.
Here's what I've tried to calculate the probability:

First, I've written the sample space ## \Omega = \{ (d_1,w),(d_1,b), (d_2,w) , (d_3,w) ,(d_3,b) \} ##. ( ## d_1,d_2,d_3 ## stand for drawers ##1,2,3 ## respectively and ##w,b## stand for White,Black )

Denote ## D_1,D_2,D_3 ## as the events of choosing drawers ##1,2,3 ## respectively. Denote ## W ## as the event of choosing a white sock.
Note that,
## D_1 = \{ (d_1,w),(d_1,b) \} ##, ## D_2 = \{ (d_2,w) \} ##, ## D_3 = \{ (d_3,w) ,(d_3,b) \} ##.
Note that
## W\cap D_1 = \{ (d_1,w) \} ##,## W\cap D_2 = \{ (d_2,w) \} ##,## W\cap D_3 = \{ (d_3,w) \} ##.

## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
Stop right there. You keep ignoring the FACT that the drawers are equally likely with probabilities ##P(D_i)=1/3##.
This is very clearly stated in the problem statement, whereas I can not follow any logic behind your numbers.
Until you get that right, there is no reason to go further.

Well, I can solve it using intuition and basic stuff, but I'm trying to solve it by first principles ( a more complicated way) for the sake of learning, not just solving.
Nobody is suggesting using intuition, we are trying to show you how to solve it by first principles. This should not be complicated, the complication is coming from your refusal to listen to what you are being told and believing that mathematics is about fancy symbols rather than understanding the concepts the symbols represent. This ends up with you writing things that are obviously incorrect like:
## P(D_1) = \frac{ |D_1|}{| \Omega |} = \frac{2}{5} ##
## P(D_2) = \frac{ |D_2|}{| \Omega |} = \frac{1}{5} ##
## P(D_3) = \frac{ |D_3|}{| \Omega |} = \frac{2}{5} ##
where it should be obvious that ## P(D_1) = P(D_2) = P(D_3) = \frac 1 3 ##.

If you want to learn, draw a tree:
• Start with a node with three branches to nodes labelled ## D_1 ## etc.
• Write the probability ## \frac 1 3 ## on each branch.
• Add a second level of branches to nodes labelled ## W ## and ## B ## .
• Write the appropriate probabilities on each branch (e.g. the branch from ## D_1 ## to ## W ## should be labelled ## \frac 1 3 ##).
• Multiply the probabilities that lead to ## W ## leaf nodes and write them next to the nodes.
• Add up the probabilities on the ## W ## leaf nodes to get ## \frac {11}{18} ##.
Now do the same with symbols:
• Write ## P(D_1) = P(D_2) = P(D_3) = \frac 1 3 ##. Notice how these correspond to the probabilities on the first level of branches in the tree.
• Write probabilities for ## P(W|D_1) ## etc. Notice how these correspond to the conditional probabilities on the second level of branches in the tree.
• Write ## P(W \cap D_1) = P(W | D_1) P(D_1) ## etc. Notice how these correspond to the probabilities on the ## W ## leaf nodes.
• Write ## P(W) = P(W \cap D_1) + P(W \cap D_2) + P(W \cap D_3) = \frac {11}{18} ##. Notice how this is the same sum as you performed on the tree.

• CGandC
The sample space could be taken to be the set of pairs (drawer chosen, colour of sock chosen). It can be taken to have 3 x 2 = 6 elements, though one of them has probability zero, so could simply be erased. After that, give these 6 points probabilities as done above. Notice that the event ‘first drawer is chosen’ ={(1,w), (1,b)}. The event “sock chosen is white” = {(1,w), (2,w), (3,w)}. To allocate probabilities to these 6 points you need to *think* about the probabilities of drawers, and about the conditional probabilities of colours given drawers.

You also could take the sample space as being a set of socks. Number the socks in each drawer, giving the black socks lower numbers than the white socks. You could even have an imaginary “number 3” sock in each drawer with only two socks. Now there are 3x3 points in your sample space. Two of them have zero probability. The probabilities of the socks in each drawer add up to 1/3. The socks with non zero probability in each drawer have equal probabilities, so either 1/3 is shared inequality parts over three socks, or over two socks.

Any rich enough sample space will do, but you will have to use the implied strong meaning of “chosen at random” in order to figure out what all those probabilities are.

What is wrong is blindly assuming that each point of your sample space has equal probability. If you thought that “first principles” means using the rule P(A) = #A / #Omega you are badly wrong. That “rule” is not a first principle but a mathematical model for the experiment “choosing a point uniformly at random from a finite set Omega”.

• CGandC
The complete sample space is
{##(D_1, b_1), (D_1, b_2), (D_1, w_1),
(D_2, w_1), (D_2, w_2),
(D_3, b_1), (D_3, w_2)##}.
It can also be specified other ways, such as {##b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1}##}, where the first index in the box number and the second is the sock number of the color in that box.
The next step is to assign probabilities to each event of picking that sock. That requires several applications of the probability of the box times the conditional probability of picking that sock given that box. They are like this:
##P(b_{1,1})=1/3*1/3=1/9##
Then you can add up all the probabilities of the white socks.

• CGandC
I would call that *a* complete sample space. Complete in the sense that it is rich enough to be able to talk about all the probabilities of all events described in the original problem. It is also minimal, in the sense that it has the smallest possible number of elementary outcomes. But it is not the only possible solution, it is not the only sensible solution. Of course, all sensible solutions end up with the same answer to the original question.

• FactChecker
Nobody is suggesting using intuition, we are trying to show you how to solve it by first principles. This should not be complicated, the complication is coming from your refusal to listen to what you are being told and believing that mathematics is about fancy symbols rather than understanding the concepts the symbols represent. This ends up with you writing things that are obviously incorrect like:

How do I learn from basics if you're letting me jump straight to deep waters by thinking of probabilities? ( not the the problem's hard, I'm talking about establishing an understanding ) first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model, then I'd start thinking about probabilities but only after I've combinatorically reasoned about the problem?
The tree helps for calculating the probabilities but I want to work bottom-up but first looking at my sample space. Thanks for the algorithm though, I find it helpful.

The sample space could be taken to be the set of pairs (drawer chosen, colour of sock chosen). It can be taken to have 3 x 2 = 6 elements, though one of them has probability zero, so could simply be erased. After that, give these 6 points probabilities as done above. Notice that the event ‘first drawer is chosen’ ={(1,w), (1,b)}. The event “sock chosen is white” = {(1,w), (2,w), (3,w)}. To allocate probabilities to these 6 points you need to *think* about the probabilities of drawers, and about the conditional probabilities of colours given drawers.

You also could take the sample space as being a set of socks. Number the socks in each drawer, giving the black socks lower numbers than the white socks. You could even have an imaginary “number 3” sock in each drawer with only two socks. Now there are 3x3 points in your sample space. Two of them have zero probability. The probabilities of the socks in each drawer add up to 1/3. The socks with non zero probability in each drawer have equal probabilities, so either 1/3 is shared inequality parts over three socks, or over two socks.

Any rich enough sample space will do, but you will have to use the implied strong meaning of “chosen at random” in order to figure out what all those probabilities are.

What is wrong is blindly assuming that each point of your sample space has equal probability. If you thought that “first principles” means using the rule P(A) = #A / #Omega you are badly wrong. That “rule” is not a first principle but a mathematical model for the experiment “choosing a point uniformly at random from a finite set Omega”.
I have thought of the first sample-space you proposed regarding the 6 elements ( but not about “sock chosen is white” = {(1,w), (2,w), (3,w)} ) and giving one element probability zero, but it seemed to me too artificial ( also in relation to the possible rich sample spaces ). And by first principles I meant defining my sample space then a probability space in correspondence to each result in the sample; it doesn't have to be a uniform probability space but that is what "feels" right for the problem.

The complete sample space is
{##(D_1, b_1), (D_1, b_2), (D_1, w_1),
(D_2, w_1), (D_2, w_2),
(D_3, b_1), (D_3, w_2)##}.
It can also be specified other ways, such as {##b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1}##}, where the first index in the box number and the second is the sock number of the color in that box.
The next step is to assign probabilities to each event of picking that sock. That requires several applications of the probability of the box times the conditional probability of picking that sock given that box. They are like this:
##P(b_{1,1})=1/3*1/3=1/9##
Then you can add up all the probabilities of the white socks.
I have thought about this sample space but I don't see how the probability for each result will be uniform given we don't differentiate between the socks (The tricky thing about the sample space imposed in the question is that we don't differentiate between the socks, and If we don't differentiate between the socks, a better sample space would be {##b_{1}, w_{1}, w_{2}, b_{3}, w_{3}##} )

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I have thought about this sample space but I don't see how the probability for each result will be uniform given we don't differentiate between the socks
The results are not uniform over all the socks. Each one has to be calculated. (They are uniform over the socks in a given drawer.)
##P(b_{1,1})=P(b_{1,2})=P(w_{1,1})=1/3*1/3=1/9##,
##P(w_{2,1})=P(w_{2,2})=1/3*1/2=1/6##, and
##P(b_{3,1})=P(w_{3,1})=1/3*1/2=1/6##
Now add up the probabilities of the white socks.
(The tricky thing about the sample space imposed in the question is that we don't differentiate between the socks, and If we don't differentiate between the socks, a better sample space would be {##b_{1}, w_{1}, w_{2}, b_{3}, w_{3}##} )
You can differentiate as much as you want to make the probability calculations simple. Then if the problem question does not differentiate, you add up the probabilities of the events that should be considered equivalent results.

The results are not uniform over all the socks. Each one has to be calculated. (They are uniform over the socks in a given drawer.)
##P(b_{1,1})=P(b_{1,2})=P(w_{1,1})=1/3*1/3=1/9##,
##P(w_{2,1})=P(w_{2,2})=1/3*1/2=1/6##, and
##P(b_{3,1})=P(w_{3,1})=1/3*1/2=1/6##
Now add up the probabilities of the white socks.

You can differentiate as much as you want to make the probability calculations simple. Then if the problem question does not differentiate, you add up the probabilities of the events that should be considered equivalent results.

Just one thing that bothers me is how do I know a-priori without the answer, If I should've differentiated between the different socks in a given drawer ( then, the sample space would be ## b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1} ## ) or if I shouldn't have ( in which case, the sample space would be ## b_{1}, w_{1}, w_{2}, b_{3}, w_{3} ## )? because it seems to me like you are differentiating between them only because you know what the answer for the probability of choosing a sock is supposed to be.

Just one thing that bothers me is how do I know a-priori without the answer, If I should've differentiated between the different socks in a given drawer ( then, the sample space would be ## b_{1,1}, b_{1,2}, w_{1,1}, w_{2,1}, w_{2,2}, b_{3,1}, w_{3,1} ## ) or if I shouldn't have ( in which case, the sample space would be ## b_{1}, w_{1}, w_{2}, b_{3}, w_{3} ## )? because it seems to me like you are differentiating between them only because you know what the answer for the probability of choosing a sock is supposed to be.
Not "supposed to be", I differentiated enough so that the individual probabilities were obvious. You have to differentiate at least to the extent that you can calculate individual probabilities and at least enough to answer the problem (often more). For that, you need to be able to interpret the problem presented so that your set of tools can be applied. Then you calculate probabilities at a fine enough resolution. Once that is done, you can combine and add probabilities to answer the particular question that the problem asks for.

How do I learn from basics if you're letting me jump straight to deep waters by thinking of probabilities? ( not the the problem's hard, I'm talking about establishing an understanding )
You are mistaken, it is you that is "trying to jump straight into the deep waters". I have shown you the way this material is taught to more than a million 14 year olds in the UK each year; IME most of them understand it.

first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model,
And this is where you are going wrong: because you are jumping straight to the sample space without considering how that sample space is constructed you can't see that the distribution is not uniform. This is what the tree is for.
then I'd start thinking about probabilities but only after I've combinatorically reasoned about the problem?
The tree is the best way to combinatorially reason about the problem.

• FactChecker
first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model,
I think this is your fundamental error. The probability distribution of the sample space is not always uniform. It is up to you to calculate the correct probabilities using a valid mathematical process.
This is a very common situation, where the process presented is done in a sequence of steps (pick a drawer, then pick a sock from the drawer). The conditional probabilities of the results of the individual steps are clear. Use those to calculate the correct probabilities of the sample space. Your assumption that the sample space has a uniform distribution is just wishful thinking.

• pbuk
You are mistaken, it is you that is "trying to jump straight into the deep waters". I have shown you the way this material is taught to more than a million 14 year olds in the UK each year; IME most of them understand it.
I've been taught this material in a school also and I understood it. But in school it is learned in a matter which doesn't make you think about what you're doing, why you're doing it and where do these steps stem from, these are research questions one intends to find answers in a place of scholarship like the academia.

You seem to be missing the important point here but I've tried explaining it to you but you keep ignoring it.
I will give you another example that can relate to the way you're viewing the current discussion: A kid knows that if an object drops down then a force of gravity is acting on it and that its weight is ## w = m \cdot g ##. One who learns physics in the the academia will look again at the problem of the object falling down but will try to answer it in a much more research and rigorous oriented way, for example, he'll make a proper free body diagram, define coordinates, define the vectors, define the assumptions, maybe look at the energy expression of the object, etc... In short, he'll try to dive deeper into the details in order to gain a better understanding of the subject which will in turn merit him later on when he will deal with more complicated problems/situations concerning the area of his study, that is - the student is intended to expand his ways of thinking about those details that concern his field of erudition.

I hope I convinced you, If so then I hope it will make you a virtue; if you still remain unconvinced and continue to be the discerning critic of everything one has to say then there won't be any agreement between us as to the idea of how one should look at problems ( which, I say, sometimes there are isn't a "single" way to look at them and the many efforts of differentiation and attempts of understanding the convoluted paths to establish a clear understanding should also be appreciated ).

I think this is your fundamental error. The probability distribution of the sample space is not always uniform. It is up to you to calculate the correct probabilities using a valid mathematical process.
This is a very common situation, where the process presented is done in a sequence of steps (pick a drawer, then pick a sock from the drawer). The conditional probabilities of the results of the individual steps are clear. Use those to calculate the correct probabilities of the sample space. Your assumption that the sample space has a uniform distribution is just wishful thinking.

Thanks, I think that's enough for now. I'll try to unthink of this problem for the mean-time and approach to it later; and if that was not helpful then I'll discuss with my professor.

I often like to use extreme examples for "sanity checks" of my thinking.
Suppose you had 10 drawers. Nine have one black sock each and the tenth has a million white socks. You pick one drawer out of the ten with uniform probability and then pick a sock from that drawer with uniform probability. Even though the vast majority of socks are white, their probabilities add up to only 1/10.

• phinds
I often like to use extreme examples for "sanity checks" of my thinking.
Yep. Often clarifies things quite nicely and usually with very little effort.

How do I learn from basics if you're letting me jump straight to deep waters by thinking of probabilities? ( not the the problem's hard, I'm talking about establishing an understanding ) first I need to think of my sample space, then I would think what probability model I'd like on the elements of the sample space, in my case in is uniform model, then I'd start thinking about probabilities but only after I've combinatorically reasoned about the problem?
I don't think you can achieve what you hope to with this problem. As others have pointed out, for the sample space you constructed, the probabilities are not uniform, so you can't simply count to calculate the probabilities. You have to calculate them using different reasoning.

You might want to consider a really simple case. You toss two unfair coins. The sample space is {HH, HT, TH, TT}. How are you going to reason using combinatorics to derive the probability for each outcome? You can't. You need some other method and use more information to assign probabilities.

While I can understand your desire to solve the problem from first principles, I don't think it's particularly useful here. It would be similar to trying to find the derivative of ##\sin x^2## using the limit definition of the derivative rather than making use of the chain rule.

• DrClaude, CGandC, pbuk and 1 other person