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

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CEO
Joined: 15 Aug 2003
Posts: 3452

Kudos [?]: 928 [0], given: 781

3) There are 12 white balls and 12 black balls that are [#permalink]

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12 Oct 2003, 21:13
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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.

Kudos [?]: 928 [0], given: 781

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 311 [0], given: 0

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13 Oct 2003, 01:03
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

Kudos [?]: 311 [0], given: 0

CEO
Joined: 15 Aug 2003
Posts: 3452

Kudos [?]: 928 [0], given: 781

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13 Oct 2003, 01:10
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

Kudos [?]: 928 [0], given: 781

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 311 [0], given: 0

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13 Oct 2003, 01:39
I think yes, for the basic formula you employ is affected by the fact that 3 black ones are out.

Kudos [?]: 311 [0], given: 0

13 Oct 2003, 01:39
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# 3) There are 12 white balls and 12 black balls that are

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