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# There are 5 different coloured balls in a bag. A ball is chosen and re

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Intern
Joined: 13 Nov 2010
Posts: 2

Kudos [?]: 1 [0], given: 2

There are 5 different coloured balls in a bag. A ball is chosen and re [#permalink]

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13 Nov 2010, 10:04
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There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

all the balls chosen are different colours?
2 of the balls are the same colour?
3 of the balls are the same colour?
all the balls are the same colour?

Working out the all same and all different is not too hard but is there a quick way to do the other parts without drawing the 625 branch tree diagram?

Kudos [?]: 1 [0], given: 2

Manager
Joined: 16 Oct 2010
Posts: 86

Kudos [?]: 70 [0], given: 3

Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 700 Q49 V35
WE: Information Technology (Investment Banking)
Re: There are 5 different coloured balls in a bag. A ball is chosen and re [#permalink]

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14 Nov 2010, 00:12
greed1 wrote:
Hi I was wondering if anyone knows how to work this question out:

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

all the balls chosen are different colours?
2 of the balls are the same colour?
3 of the balls are the same colour?
all the balls are the same colour?

Working out the all same and all different is not too hard but is there a quick way to do the other parts without drawing the 625 branch tree diagram?

I would like to give it a try.
1) All of the balls cosen are of different color
P = 5/5 * 4/5 * 3/5 * 2 /5 = 24/125

2) p = 5/5 * 1/5 * 4/5 * 3/5 * 4C2 = 72/125

3) p = 5/5 * 1/5 * 1/5 * 4/5 * 4C3 = 16/125

4) p = 5/5 * 1/5 * 1/5 * 1/5 = 1/125

Kudos [?]: 70 [0], given: 3

Intern
Joined: 13 Nov 2010
Posts: 2

Kudos [?]: 1 [0], given: 2

Re: There are 5 different coloured balls in a bag. A ball is chosen and re [#permalink]

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14 Nov 2010, 02:28
1
This post was
BOOKMARKED
syog wrote:
greed1 wrote:
Hi I was wondering if anyone knows how to work this question out:

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

all the balls chosen are different colours?
2 of the balls are the same colour?
3 of the balls are the same colour?
all the balls are the same colour?

Working out the all same and all different is not too hard but is there a quick way to do the other parts without drawing the 625 branch tree diagram?

I would like to give it a try.
1) All of the balls cosen are of different color
P = 5/5 * 4/5 * 3/5 * 2 /5 = 24/125

2) p = 5/5 * 1/5 * 4/5 * 3/5 * 4C2 = 72/125

3) p = 5/5 * 1/5 * 1/5 * 4/5 * 4C3 = 16/125

4) p = 5/5 * 1/5 * 1/5 * 1/5 = 1/125

The right method. Your answers add up to 113/125, so what would account for the remaining 12/125?

Kudos [?]: 1 [0], given: 2

Retired Moderator
Joined: 02 Sep 2010
Posts: 793

Kudos [?]: 1184 [1], given: 25

Location: London
Re: There are 5 different coloured balls in a bag. A ball is chosen and re [#permalink]

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14 Nov 2010, 03:15
1
KUDOS
greed1 wrote:
Hi I was wondering if anyone knows how to work this question out:

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

all the balls chosen are different colours?
2 of the balls are the same colour?
3 of the balls are the same colour?
all the balls are the same colour?

Working out the all same and all different is not too hard but is there a quick way to do the other parts without drawing the 625 branch tree diagram?

1) All balls of different colors --> (5/5) * (4/5) * (3/5) * (2/5) = (24/125)

2) 2 balls of the same color --> ((pick the color to be repeated twice) * (pick the other 2 color) * (number of permutations))/(Total permutations) = (5 * C(4,2) * (4!/2!))/(5^4) = (72/125)

3) 3 balls of the same color --> ((pick the common color) * (pick the other color) * (number of permutations))/(total permutations) = (5 * C(4,1) * (4!/3!))/5^4 = (16/125)

4) All same color --> Only 1 such combination possible for each color, there are 5 colors, so total probability = 5/5^4 = (1/125)

5) Two balls of one color and two of another color --> ((choose 2 colors) * (permute))/(total permutations) = C(5,2) * (4!/2!2!) / 5^4 = (12/125)

Total probability for all cases = 1
_________________

Kudos [?]: 1184 [1], given: 25

Manager
Joined: 16 Oct 2010
Posts: 86

Kudos [?]: 70 [1], given: 3

Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 700 Q49 V35
WE: Information Technology (Investment Banking)
Re: There are 5 different coloured balls in a bag. A ball is chosen and re [#permalink]

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14 Nov 2010, 03:28
1
KUDOS
greed1 wrote:
syog wrote:
greed1 wrote:
Hi I was wondering if anyone knows how to work this question out:

There are 5 different coloured balls in a bag. A ball is chosen and replaced 4 times. What is the probability that:

all the balls chosen are different colours?
2 of the balls are the same colour?
3 of the balls are the same colour?
all the balls are the same colour?

Working out the all same and all different is not too hard but is there a quick way to do the other parts without drawing the 625 branch tree diagram?

I would like to give it a try.
1) All of the balls cosen are of different color
P = 5/5 * 4/5 * 3/5 * 2 /5 = 24/125

2) p = 5/5 * 1/5 * 4/5 * 3/5 * 4C2 = 72/125

3) p = 5/5 * 1/5 * 1/5 * 4/5 * 4C3 = 16/125

4) p = 5/5 * 1/5 * 1/5 * 1/5 = 1/125

The right method. Your answers add up to 113/125, so what would account for the remaining 12/125?

as suggested above, the remaining 12/125 is the probability of getting 2 balls of one colour and another 2 balls of another one colur.

p = 5/5 * 1/5 * 4/5 * 1/5 * 4C2/2 = 12/125

When we say probability of getting 2 balls of one colour, it inherently means that remaining balls are of different colours.

Kudos [?]: 70 [1], given: 3

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Joined: 09 Sep 2013
Posts: 16787

Kudos [?]: 273 [0], given: 0

Re: There are 5 different coloured balls in a bag. A ball is chosen and re [#permalink]

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01 Oct 2017, 06:09
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Re: There are 5 different coloured balls in a bag. A ball is chosen and re   [#permalink] 01 Oct 2017, 06:09
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