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MihirBathia

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31 Jan 2024, 02:00

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Wasted 15 minutes and now feeling stupid as hell

HarshaBujji

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08 Nov 2024, 10:56

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Bunuel

A drawer contains 5 pairs of white socks, 3 pairs of black socks, and 2 pairs of grey socks. If four individual socks are randomly chosen without replacement, what is the probability of getting at least two socks of the same color?

A. \(\frac{1}{5}\)

B. \(\frac{2}{5}\)

C. \(1\)

D. \(\frac{3}{5}\)

E. \(\frac{4}{5}\)

Pegion Hole principle : The pigeonhole principle is a mathematical theorem that states that if more items are put into fewer containers, then at least one container must contain more than one item.

In the same lines here there are only 3 varieties, We need to choose 4 so its obvious that one of the variety will be repeated.

Hence Probability is 1.

Chandan1991

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04 Jul 2025, 23:52

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I did not quite understand the solution. It says individual socks, not 4 individual pairs of socks. Individual socks may mean 20 socks of different colours i.e 10 white, 6 black and 04 grey

Bunuel

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05 Jul 2025, 01:29

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Chandan1991

I did not quite understand the solution. It says individual socks, not 4 individual pairs of socks. Individual socks may mean 20 socks of different colours i.e 10 white, 6 black and 04 grey

The socks are individual: 10 white, 6 black, 4 grey, total 20. But with only 3 colors, choosing 4 socks forces at least one color to repeat. So the probability is 1. Please review the question and solution more carefully.

Paattaa

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18 Sep 2025, 22:27

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This is a great question that’s helpful for learning.

TheSyntaxerror101

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23 Dec 2025, 20:52

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I like the solution - it’s helpful.

SachinNayak

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15 Apr 2026, 04:04

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Hi Bunuel
I had this doubt about complimentry rule.
So usally Ive seen when the question asks atleast for example when there are only red and blue balls in a bowl. What is the probability of getting atleast 1 red ball when three balls are drawn randomly from the bowl?Given that bowl has 5 blue and 4 red balls.
we would do 1- no red ball = 1- (5/9*4/8*3/7). Here my understanding is when it says atleast 1 ball it could be 1 red ball or 2 red ball or all of them being red(3 red balls). So I thought P(0)+P(1)+P(2)+P(3)=1 P(0) means 0 red balls in 3 draws,P(1) means only 1 red ball so on.Instead of finding for all we just do 1-P(0) (which is no red ball.)Now in this ques it said P(atleast two socks of same colour) which would be 1- (P(0)+P(1)). P(0) is 0 as its not possible but why isn't P(1) there. I think i got the lamguage wrong here? It said atleast two socks of same colour and for P(1) I thought 1 sock of same colour which I thought to be WWBG,BBWG,GGWB so after 3 socks are drawn 1 sock of same colour that has been already drawn?. When I re-read it I stared reading it as P(Atleast two socks of same colour) = 1- P(no two socks of same colour) and not 1 - (P(0)+P(1)). Its the socks word understanding I went wrong?Let me know

Thanks

Bunuel

WWWWWWWWWW - BBBBBB - GGGG

We need to find the probability of the following scenarios:

1. Two socks of one color and one sock each of the other two colors (XXYZ)
WWBG
BBWG
GGWB

2. Three socks of one color and one sock of another color (XXXY)
WWWB
WWWG
BBBW
BBBG
GGGW
GGGB

3. Two socks of each of two colors (XXYY)
WWBB
WWGG
BBGG

4. Four socks of the same color (XXXX)
WWWW
BBBB
GGGG

Calculating the probabilities of each scenario separately would be tedious. Instead, we can find the probability of the complementary event (i.e., selecting four socks with no two socks of the same color) and subtract it from 1. However, it is not possible to draw four socks with no two socks of the same color, since we only have three colors. Therefore, the probability of the complementary event is 0.

Thus, the probability of getting at least two socks of the same color is:

P(at least two of the same color) = 1 - P(complementary event) = 1 - 0 = 1.

Bunuel

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15 Apr 2026, 04:23

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SachinNayak

Yes, your second reading is the correct one. Here “at least two socks of the same color” means at least one matching color appears, so the complement is “no two socks are the same color,” not “exactly one sock of some color.”

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