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I'm selecting 164 items from a bag of 394. What is the probability that exactly 162 of these items will be green if there are 264 green items in the bag?

I calculated this by multiplying the probability of selecting a green object on the first try (264/394) by the probability of selecting a green object on the second try (263/393) by the probability of selecting a green object on the third try (262/392) and so on and so forth 162 times. Then I multiply this product by the probability that the last two draw are NOT green (130/232)*(129/231).

This results in a crazy small number. I just don't think this makes sense because if I follow this logic and calculate the probability of selecting 110 green objects and 54 non-green objects, which would result in a sample that matches the distribution of the population, I also get a very small probability.

mdrishan
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  • Your method appears to specify the order in which the various colors are chosen...you have to consider all possible orders. Doesn't matter much, the probability is still extremely low. – lulu Dec 16 '19 at 22:09
  • The number of ways to draw precisely this combination of colors is $\binom {264}{162}\times \binom {130}2$. The number of ways to draw $164$ items out of $394$ with no restriction on colors is... – lulu Dec 16 '19 at 22:20

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