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111
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avatar+893 

Can anyone solve this?

 Mar 13, 2019
 #1
avatar+5655 
0

incorrect answer deleted

.
 Mar 14, 2019
edited by Rom  Mar 14, 2019
edited by Rom  Mar 14, 2019
edited by Rom  Mar 14, 2019
 #2
avatar+893 
0

The answer actually was 8/15. Do you know how?

dgfgrafgdfge111  Mar 14, 2019
 #3
avatar+5655 
+3

sim shows 8/15 is correct.

 

Ok, it's not that big a deal.

 

You choose the first sock with probability 1 and the second unmatched sock with probability 4/5

 

For clarity let's say you chose (r,w) as the first pair.  We are left with (r,w,b,b)

For the third sock you can choose, either r or w, and be forced to choose b as the 4th, or

You can choose b, and then be forced to choose r or w as the 4th.  This is

 

2/4 x 2/3 + 2/4 x 2/3 = 2/3

 

I.e. there is a 2/3 probability you will choose a non-matching pair for the 3rd and 4th socks.

 

P[non-matching pair] = 4/5 x 2/3 = 8/15

Rom  Mar 14, 2019
edited by Rom  Mar 14, 2019
edited by Rom  Mar 14, 2019
 #4
avatar+893 
+1

Ah, I see. That's a nice solution. Thank you!!!

dgfgrafgdfge111  Mar 14, 2019

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