+0

# 2016 NS 21

0
235
4
+935

Can anyone solve this?

Mar 13, 2019

#1
+6196
0

.
Mar 14, 2019
edited by Rom  Mar 14, 2019
edited by Rom  Mar 14, 2019
edited by Rom  Mar 14, 2019
#2
+935
0

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

dgfgrafgdfge111  Mar 14, 2019
#3
+6196
+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
+935
+1

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

dgfgrafgdfge111  Mar 14, 2019