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If I had 9 coat hangers (for example: 4 green, 3 blue and 2 red), what would be the probability of the first four being [green, green, blue, blue], if I just chuck the coat hangers into the wardrobe at random?

I have a feeling this is a permutations question, but I don't have that much experience in statistics to know how to solve it properly.

Could anyone help me solve this problem, or at least steer me in the right direction?

Canada
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  • Do you count outcome `GGBB` as distinct from `BBGG` and `BGBG`~, etc. ? And please tell us what you have tried. – BruceET Jan 31 '22 at 01:34
  • @BruceET Well, the question is what would be the probability of the first two being green and the second two blue. So the important thing would be that the first four are GGBB. I've tried using the permutations equation and ended up with 5!/9! as the answer. Could you tell me where I went wrong? – Canada Jan 31 '22 at 02:01
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    What is the probability that the first is green. Then one hanger is taken, so what is the probability the second is also green, etc. // General multiplication rule. – BruceET Jan 31 '22 at 02:04
  • Aaaah, I just crossed the first four off as only one coat hanger could go in each of those places, and afterwards I was left with five random coat hangers, so that's where I got the 5! from, and I divided that with 9! which would be the total number of combinations in the wardrobe. – Canada Jan 31 '22 at 02:10
  • I suppose you've figured out by now that $5!/9! \ne 0.0238$ or anywhere near. – BruceET Jan 31 '22 at 02:13
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    Yeah, I have, thank you for the swift answer. Cheers! – Canada Jan 31 '22 at 02:15

2 Answers2

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Think about what's the probability of first hanger being green. There are 9 hangers in total and there are 4 green so it would be $\frac{4}{9}$ for the second time we have 3 green hanger and 8 total hangers so it would be $\frac{3}{8}$. You can apply same logic for the blue ones and we get $\frac{4}{9} \times \frac{3}{8} \times \frac{3}{7} \times \frac{2}{6} = \frac{1}{42}$.

This would be the answer if the order is specifically Green, Green, Blue and Blue if we are talking about any combination of 2 green and 2 blue then we need to multiply it with the total permutations of GGBB which would be $\frac{4!}{2 \times 2} = 6$ So the final answer would be $\frac{1}{42} \times 6 = \frac{1}{7}$

Altaïr Ibn-La'Ahad
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Comment: Until you show what you have tried, I will not discuss combinatorial answers. But if order is important, then the following simulation in R gives a good approximation of the correct answer: $0.0238 \pm 0.0020.$

 set.seed(2022)
 pop = c(1,1,1,1,2,2,2,3,3)
 match = replicate(10^6, 
            sum(sample(pop,4)==c(1,1,2,2)))
 mean(match==4)
 [1] 0.023761       # approx P(GGBB)
 2*sd(match)/1000
 [1] 0.001968896    # approx margin of sim error
BruceET
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  • Yes, I think this is it. It would appear that I forgot to mention that the order is important and instead just talked about permutations, my bad. – Canada Jan 31 '22 at 02:06