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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which

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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which  [#permalink]

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New post 03 Jul 2017, 02:13
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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which are dispensed at random. If the first three gum balls dispensed are black, what is the probability that the next two gum balls dispensed will also be black?

(A) 6/35
(B) 4/15
(C) 1/3
(D) 3/7
(E) 1/2

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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which  [#permalink]

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New post 03 Jul 2017, 02:41
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Given data : 24 gum balls - 12 white and 12 black
3 black gum balls have already been chosen

To find : Probability that next 2 gum balls are also black in color

Denominator(2 gum balls chosen from 21 remaining balls) \(21c2 = \frac{21*20}{2} = 210\)
Numerator(Balls picked are black) \(9c2 = \frac{9*8}{2} = 36\)

Probability that next 2 gum balls are also black in color(of remaining 9 black gum balls) = \(\frac{36}{210} = \frac{6}{35}\)(Option A)
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Re: A gum ball dispenser has 24 gum balls: 12 white and 12 black, which  [#permalink]

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New post 03 Jul 2017, 10:51
Since already 3 black balls have been chosen, so probability to choose 2 more black balls out of remaining 9 black balls = 9C2 = 9*4

Probability to choose 2 balls from 21 balls = 21C2 = 21*10

probability that next two balls will be black = 9*4 / 21*10 = 6/35
Ans:A
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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which  [#permalink]

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New post 04 Jul 2017, 09:32
Bunuel wrote:
A gum ball dispenser has 24 gum balls: 12 white and 12 black, which are dispensed at random. If the first three gum balls dispensed are black, what is the probability that the next two gum balls dispensed will also be black?

(A) 6/35
(B) 4/15
(C) 1/3
(D) 3/7
(E) 1/2

First three gumballs dispensed were black. That info yields two changes:

1. The total number of gumballs decreases from 24 to 21; and

2. The number of black gumballs decreases from 12 to 9

What is the probability that the next two gumballs dispensed will be black?

First dispensation, probability is \(\frac{9}{21}\).That leaves 8 black and 20 total gumballs.

Second dispensation: probability is \(\frac{8}{20}\)

\(\frac{9}{21}\) * \(\frac{8}{20}\) =

\(\frac{3}{7}\) * \(\frac{2}{5}\) = \(\frac{6}{35}\)

Answer A
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Re: A gum ball dispenser has 24 gum balls: 12 white and 12 black, which  [#permalink]

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New post 09 Jul 2017, 10:26
Earlier - 24 is 12 w and 12 b
if 3 black balls dispensed:
the ration is:
total 21
12 white
9 black
probability for next two to be black is 9/21 * 8/20 = 3/7*2/5 = 6/35 ----> Option A

+1 kudos if u like post.
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Re: A gum ball dispenser has 24 gum balls: 12 white and 12 black, which  [#permalink]

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New post 09 Jul 2017, 19:23
\(\frac{9}{21} * \frac{8}{20} = \frac{3}{7} * \frac{2}{5} = \frac{6}{35}\). Ans - A.
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Re: A gum ball dispenser has 24 gum balls: 12 white and 12 black, which   [#permalink] 09 Jul 2017, 19:23
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