There are three red, three blue, and three green balls in an opaque ba

Question

There are three red, three blue, and three green balls in an opaque bag. Alan draws three balls at random without replacement, then passes the bag to Bridget, who does the same, as does Charlie. What is the probability that Charlie will draw three balls of the same color?

A. 1/56
B. 1/28
C. 1/14
D. 1/4
E. 1/3

A. 1/56
B. 1/28
C. 1/14
D. 1/4
E. 1/3

GMATBusters · Accepted Answer

As we don't know the outcomes of the Alan's and Bridget's draw. Their draws will not have any effect on the TOTAL probability of final draw of Charlie. Hence draw of A & B can be can be neglected.
Just observe Charlie's draw.

3 cases to draw same colour = 3C1=3.
Total cases = 9C3

Required probability =3/9C3 = 1/28.

WizakoBaskar · Answer

If the 3 balls that Charlie draws are all red, then Alan and Bridget should have drawn their balls from the remaining 6.

Probability that Alan draws 3 balls, none of which is red = (number of ways of drawing 3 balls from remaining 6)/(number of ways of drawing 3 from 9) 
=  \frac{6C3}{9C3}

Alan has already picked 3 balls. So, Bridget has to pick 3 balls out of the remaining 6. If none of those are to be red, Bridget has to pick 3 out of the 3 balls that are not red.

Probability that Bridget draws 3 balls, none of which is red = (number of ways of drawing 3 balls from 3 balls that are not red)/(number of ways of drawing 3 from 6)
 =  \frac{3C3}{6C3}

So, the probability that Charlie draws 3 red balls =  \frac{6C3}{9C3} * \frac{3C3}{6C3}  =  \frac{3C3}{9C3}  =  \frac{1}{84}

The 3 balls that Charlie draws should be of the same colour. So, it could be all red or all blue or all green.
So, probability = 3 *  \frac{1}{84}  =  \frac{1}{28}

Choice B.

GyMrAT · Answer

We are looking for a probability of 3 same colored balls & 6 any other colored balls in the order OOOOOOSSS, O is any other colored ball & S = Green/Red/Blue

The order can be OOOOOOGGG / OOOOOORRR / OOOOOOBBB

Hence we have required probability = (6/9)*(5/8)*(4/7)*(3/6)*(2/5)*(1/4)*(3/3)*(2/2)*(1/1)*3 = 1/28

Answer B.

Thanks,
GyM

DavidTutorexamPAL · Answer

We'll break our problem down into smaller pieces so it is easier to calculate.
This is a Logical approach.

Let's start with one color: red.
If Charlie picked only red balls, then Alan and Bridget must, together, have taken all of the green and blue balls.
That is, there are 6! ways for Alan and Bridget to pick these 6 balls followed by 3! ways for Charlie to pick the red balls.
This gives 6!3! ways for Charlie to pick 3 red balls.
Similarly, there are 6!3! ways to pick 3 green and 6!3! ways to pick 3 blue balls.
In total we have 3*6!3! ways for Charlie to pick 3 balls of the same color.
As there are a total of 9! orders to for picking all the balls, our probability is 3*6!3!/9! = (canceling out 6!) (3*3*2*1)/(9*8*7) = 1/(4*7) = 1/28

(B) is our answer.

Kinshook · Answer

Given: 
1. There are three red, three blue, and three green balls in an opaque bag. 
2. Alan draws three balls at random without replacement, then passes the bag to Bridget, who does the same, as does Charlie.

Asked: What is the probability that Charlie will draw three balls of the same color?

Red -3
Blue - 3
Green -3

Total no of ways = 9!/3!3!3! = 1680

# of favourable ways = 3 * 6C3 * 3C3 = 3*20 = 60.

Probability = 60/1680 = 1/28

IMO B

Archit3110 · Answer

total possible cases ; 9c3 ; 84
and P for charlie ; 3c1 ; 3
P for charlie ; 3/84 ; 1/28
IMO B

dave13 · Answer

VeritasKarishma i used method (1- all non favorable results) and focused only on Charlie. Case 1. Charlie draws two of the same color and one of another \frac{3}{9}*\frac{2}{8}*\frac{3}{7} = \frac{1}{28} so since i can have 6 such events: (2R +1G, 2R+1B, 2B+1G, 2B+1R, 2G+1R, 2G+1B ) multiply 1/28 by 6 = \frac{3}{14} Case 2: Charlie draws 3 balls of different colors (R *G * B) \frac{3}{9}*\frac{3}{8}*\frac{3}{7} = \frac{3}{56} So total number of unfavorable outcomes for Charlie is \frac{3}{14}+\frac{3}{56} = \frac{15}{56} so i get this answer 1- \frac{15}{56} = \frac{41}{56 }

can you pls let me know where am I making mistake ?

KarishmaB · Answer

Yes, whether Alan and Bridget pick 6 balls and Charlie is left with 3 or Charlie picks 3 balls and Alan and Bridget are left with 6 is the same. Hence, focus on Charlie is fine.

But it is easier to calculate the direct probability.

In the first pick, Charlie can pick any colour ball. Probability = 1
In the second pick, Charlie must pick the same colour as picked previously. Probability = 2/8
In the third pick, Charlie must pick again the same colour as picked 2 times before. Probability = 1/7
Total Probability = 1*(2/8)*(1/7) = 1/28

Answer (B)

If you do go the (1 - Non favourable) way,

Charlie pick 2 balls of same colour: First pick any colour, second pick same colour as before, third pick a different colour. Say RRB. We multiply by 3!/2! to get arrangements of RRB.
Probability = 1*(2/8)*(6/7) * 3!/2! = 18/28

Charlie picks all 3 balls of diff colours: First pick any colour, second pick another colour, third pick the third colour. 
Probability = 1*(6/8)*(3/7) = 9/28
Total Probability = 27/28

Required probability = 1 - 27/28 = 1/28

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There are three red, three blue, and three green balls in an opaque ba

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