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Probability - selecting socks

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Probability - selecting socks

by Brent@GMATPrepNow » Sat Jan 31, 2009 9:14 am
A drawer contains 5 black socks, 3 green socks and 2 yellow socks. If four socks are removed from the drawer at random, what is the probability that at least two black socks are selected?
A) 15/21
B) 5/7
C) 31/42
D) 3/4
E) 5/6

Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
Last edited by Brent@GMATPrepNow on Sat Jan 31, 2009 12:17 pm, edited 2 times in total.
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by aroon7 » Sat Jan 31, 2009 9:19 am
Sorry Bernt. We expect to have atleat two black socks in our hand or in the drawer?
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by aroon7 » Sat Jan 31, 2009 9:36 am
Here is what i did:


selecting 2 black socks = 5C2
selecting remaining socks = 8C2 (2 socks already selected out of 10. so 8 socks remain and we need to pick 2 out of these 8)

prob = 5C2 * 8C2 / 10C4 = 280/210 :oops: this is greater than 1
some one please help me...
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by Brent@GMATPrepNow » Sat Jan 31, 2009 9:44 am
aroon7 wrote:Sorry Bernt. We expect to have atleat two black socks in our hand or in the drawer?
Sorry, that question was poorly worded.
We want the probability of selecting at least 2 black socks.
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by dmateer25 » Sat Jan 31, 2009 10:51 am
I am having trouble with this one:

Here is what I come up with

10C4 = 210

4 blacks, no others (5C4)*(5C0) = 5

3 blacks, 1 other (5C3)*(5C1) = 10 * 5 = 50

2 blacks, 2 others (5C2)*(5C2) = 10 * 10 = 100

5 + 50 + 100 = 155

155/210 = 31/42
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by Brent@GMATPrepNow » Sat Jan 31, 2009 12:13 pm
dmateer25 wrote:I am having trouble with this one:

Here is what I come up with

10C4 = 210

4 blacks, no others (5C4)*(5C0) = 5

3 blacks, 1 other (5C3)*(5C1) = 10 * 5 = 50

2 blacks, 2 others (5C2)*(5C2) = 10 * 10 = 100

5 + 50 + 100 = 155

155/210 = 31/42
Sorry, dmateer25. This is perfectly correct. The answer is 31/42.
Looks like I need to double-check my work.

I calculated 5C4 to be equal to 25 - wow!! :oops:

I've changed the original answer choice to avoid confusing others.
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by awilhelm » Sat Jan 31, 2009 1:53 pm
Hi,

Could anyone explain how to arrive at the answer without using the combination formula? Maybe using the "1 - P(not choosing any black socks)" approach? Thanks
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by umaa » Sat Jan 31, 2009 2:28 pm
Guys, won't you consider the number of arrangements?
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by DanaJ » Sat Jan 31, 2009 2:30 pm
umaa: no, since you are not interested in the which order you pick the socks. You only need 2 black, doesn't matter if one is the third and the other one is the fourth or if one is the first and the other one is third and so on...
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by umaa » Sat Jan 31, 2009 2:38 pm
got it. thanks..
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by dmateer25 » Sat Jan 31, 2009 2:51 pm
awilhelm wrote:Hi,

Could anyone explain how to arrive at the answer without using the combination formula? Maybe using the "1 - P(not choosing any black socks)" approach? Thanks
Ok, I will try:

Here are the combinations that would have less than 2 black socks.

X will be one of the other socks and be will be a black sock

XXXX
Probability of choosing no black socks would be: 1/2 * 4/9 * 3/8 * 2/7 = 1/42

BXXX
Probability of choosing 1 black and 3 other socks would be: 1/2 * 5/9 * 1/2 * 3/7 = 5/84

XBXX
Same as above: 5/84

XXBX
Same as above: 5/84

XXXB
Same as above: 5/84

We need to sum all of these up:
5/84 + 5/84 + 5/84 +5/84 +1/42 = 22/84 = 11/42

Now to find the probability of at least 2 black socks we would take 1 – 11/42 = 31/42
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by awilhelm » Sat Jan 31, 2009 3:07 pm
Thanks!
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