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Mathematics 14 Online
OpenStudy (adamaero):

Show your math to get MEDAL. Probability - If you pull out two socks (then a 3rd) from a sock drawer with two different kinds of socks, what is the chance of getting a complete pair on the 3rd random pick?

OpenStudy (perl):

probability = number of favorable / number of total

OpenStudy (adamaero):

If I stick my hand into my sock drawer, on the 2nd pick I will either have a pair or two different socks; on the 3rd pick, there is a 100% chance I'll make a pair. I'm wondering how to represent that mathematically.

OpenStudy (adamaero):

AB BA already picked A || B 3rd either A or B So end will be either ABA or BAB or BAA or ABB

OpenStudy (adamaero):

I guess I could already have a pair--tho unlikely, since I do this every day. I rarely get a pair on the first try.

OpenStudy (misty1212):

lol

OpenStudy (perl):

this is called pigeon hole principle

OpenStudy (misty1212):

if there are only two colors, i would say \(100\%\) for absolute sure you will have a pair on the third pick

OpenStudy (perl):

you have three pigeons (socks) and 2 pigeon holes (2 colors)

OpenStudy (misty1212):

pigeons smidgeons, it is just obvious

OpenStudy (perl):

a pigeon hole is going to have more than one pigeon :)

OpenStudy (adamaero):

ahh, but I want to see the math misty1212

OpenStudy (perl):

Imagine you have white socks and black socks mixed. You pick 3 socks. you are guaranteed to pick 2 socks of the same color on the third draw. This is mathematically called the piegon hole principle.

OpenStudy (perl):

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OpenStudy (adamaero):

"for natural numbers k and m, if n = km + 1 objects are distributed among m sets, then the pigeonhole principle asserts that one of the sets will contain at least k + 1 objects.[3] For arbitrary n and m this generalizes to k + 1 = ⌊(n - 1)/m⌋ + 1, where ⌊...⌋ is the floor function." haaaaaa, wikipedia is awesome: http://en.wikipedia.org/wiki/Pigeonhole_principle#Sock-picking

OpenStudy (perl):

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