- #1

CGandC

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**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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