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3) There are 12 white balls and 12 black balls that are

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CEO
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3) There are 12 white balls and 12 black balls that are [#permalink] New post 12 Oct 2003, 20:13
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3) There are 12 white balls and 12 black balls that are dispersed
randomly. If the first 3 balls are black, what is the probability
that fourth is also black.
SVP
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Joined: 03 Feb 2003
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 [#permalink] New post 13 Oct 2003, 00:03
P (4 black) = P (3 black) * P (1b/3b)

P (4 black) = 12C4/24C4
P (3 black) = 12C3/24C3

P (1b/3b)=(12C4/24C4)/(12C3/24C3)=9/21=3/7
CEO
CEO
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Posts: 3470
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Kudos [?]: 649 [0], given: 781

 [#permalink] New post 13 Oct 2003, 00:10
stolyar wrote:
P (4 black) = P (3 black) * P (1b/3b)

P (4 black) = 12C4/24C4
P (3 black) = 12C3/24C3

P (1b/3b)=(12C4/24C4)/(12C3/24C3)=9/21=3/7


Stolyar

Known that first three balls are black, so now we have 12 white and 9

black.

Then you pick the next ball, it being black will have a probability of

9/21 = 3/7

That begs the question, is my thinking right?

thanks
praetorian
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 [#permalink] New post 13 Oct 2003, 00:39
I think yes, for the basic formula you employ is affected by the fact that 3 black ones are out.
  [#permalink] 13 Oct 2003, 00:39
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