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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.

The question is fairly simple to understand but is there any other way to solve it without counting all the different possibilities of the first sample ?

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  • $\begingroup$ Unless you have a typo, then it's 0, as there will be 6 balls left after two draws. $\endgroup$ – Stefan4024 Oct 2 '16 at 9:54
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    $\begingroup$ @Stefan4024 by latter they mean the second sample $\endgroup$ – Mike Harb Oct 2 '16 at 10:05
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The question is fairly simple to understand but is there any other way to solve it without counting all the different possibilities of the first sample ?

Yes.   Ignore the first draw as it is a red herring.

Consider instead of balls you have a deck of cards with coloured balls drawn on them.   Shuffle the deck, cut the first six cards from the top and put them on the bottom of the deck, then draw eight cards and look at them.   Alternatively just shuffle the deck and immediately draw eight cards and look at them.   In terms of probability of what colours those eight cards contain, does it matter where in the deck they were drawn?


What is the probability that a sample of $8$ from the $20$ balls contains $4$ from the $10$ blue, $1$ from the $5$ yellow, and $3$ from the $5$ red balls?

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  • $\begingroup$ Yeah it becomes fairly simple with that in mind, cards always help :) $\endgroup$ – Mike Harb Oct 2 '16 at 14:07

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