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I've been trying to solve this exercise for a few hours, I can't seem to get the right answer. Can someone share the intuition or twist behind this problem? I'll perhaps try to take it from there.

An urn contains 10 blue balls , 5 yellow balls and 5 red balls. Six balls are drawn from the urn and discarded, their color unnoticed. A second sample of 8 balls is then drawn without replacement. What is the probability that the latter contains 4 blue balls, 1 yellow ball and 3 red balls.

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"Six balls are drawn from the urn and discarded, their color unnoticed."

This part is only there to confuse you. Obviously each individual ball have the same odds of being part of the second sample, so this problem is exactly the same as simply drawing 8 balls from the urn when the 20 balls are still in it:

An urn contains 10 blue balls , 5 yellow balls and 5 red balls. A sample of 8 balls is drawn without replacement. What is the probability that it contains 4 blue balls, 1 yellow ball and 3 red balls ?

Can you take on from there ?

Edit: to go a bit further, taking comment into account. The total number of possible draws is $\binom{20}{8}$

You just have to calculate the number of favourable draws: $\binom{10}{4}\binom{5}{1}\binom{5}{3}$ and take the quotient.

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  • $\begingroup$ Could you please elaborate a bit more? I know the probabilities of obtaining any of the balls but I'm still lost in how to apply the combinations, I can tell that the denominator should be $20C8$ I mean what is the methodology to solve such problems I have a huge set of problems I'd like to solve. $\endgroup$
    – hallaksec
    Jun 27, 2017 at 16:18
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    $\begingroup$ @hallaksec For the numerator, choose four of the ten blue balls, one of the five yellow balls, and three of the five red balls. $\endgroup$
    – N. F. Taussig
    Jun 27, 2017 at 23:18
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    $\begingroup$ There are ten blue balls, so the first term in the numerator should be $\binom{10}{4}$. $\endgroup$
    – N. F. Taussig
    Jun 28, 2017 at 10:08
  • $\begingroup$ @N.F.Taussig : thanks, corrected. $\endgroup$
    – Evargalo
    Jun 29, 2017 at 8:42

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