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

Question

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

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

pushpitkc · Answer

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)

nipunjain14 · Answer

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

generis · Answer

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

sonikavadhera · Answer

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.

TimeTraveller · Answer

\frac{9}{21} * \frac{8}{20} = \frac{3}{7} * \frac{2}{5} = \frac{6}{35} . Ans - A.

MathRevolution · Answer

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

gmatpro15 · Answer

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

Abhishek009 · Answer

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)

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards