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505-555 Level|   Probability|      
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Bunuel
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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 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
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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 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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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 03 Jul 2017, 10:51
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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 04 Jul 2017, 09:32
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Bunuel
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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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 09 Jul 2017, 10:26
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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

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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 09 Jul 2017, 19:23
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\(\frac{9}{21} * \frac{8}{20} = \frac{3}{7} * \frac{2}{5} = \frac{6}{35}\). Ans - A.
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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 14 May 2021, 01:24
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Total balls: 24
White: 12
Black: 12

3 balls are already black in start:

Total balls: 24 - 3 = 21
White: 12
Black: 12 - 3 = 9

Next one: P(Black) = \(\frac{9}{21}\)

Total balls: 21 - 1 = 20
White: 12
Black: 9 - 1 = 8

Last one: P(Black) = \(\frac{8}{20}\)

Probability of next two black: \(\frac{9}{21} * \frac{8}{20} = \frac{6}{35}\)

Answer A
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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 14 Aug 2022, 04:10
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total balls = 12+12
12 white balls
12 black balls

since 3 balls are already drawn
total possibilities are 21

P(GETTING BLACK BALL ON 4TH AND 5TH DRAW) = 9/21 x 8/20 = 6/35

since there will be 9 black balls left and also after getting a black in 4th draw 8 black balls will be left and total balls 20
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A gum ball dispenser has 24 gum balls: 12 white and 12 black, which 18 Jan 2024, 08:08
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Bunuel
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
No of Balls after 3 black gums are : 12 w & 9 b

1st Black ball : \(\frac{9}{21}\)
2st Black ball : \(\frac{8}{20}\)

Required Probability : \(\frac{9}{21}*\frac{8}{20} = \frac{3}{7}*\frac{2}{5} =\frac{6}{35} \), Answer will be (A)
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