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# A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls

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Intern
Joined: 13 Feb 2012
Posts: 15
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A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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17 Mar 2012, 06:28
11
00:00

Difficulty:

55% (hard)

Question Stats:

68% (02:52) correct 32% (02:49) wrong based on 240 sessions

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A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?

A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1
Manager
Joined: 22 Feb 2012
Posts: 85
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GMAT 2: 670 Q42 V40
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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17 Mar 2012, 06:45
Not sure if this is the right method but here is what i did...

ok so jar contains 20 red 20 blue 6 yellow and 4 pink..

Adding the number of balls we get 50 total..

Now its a counting + probability exercise..

2 stages exist: Pick first ball then second ball..

Denominator for first ball = 50 since there are 50 balls
denominator for second ball = 49 since only 49 would remain

for each type ball i constructed probability of getting 2 of same.. I used a counting method.. first ball 20 chances blue and 2nd ball 19 chances blue and so on and so forth

Blue = (20*19)/(50*49) rounded to 400/2500
Red = (20*19)/(50*49) rounded to 400/2500
Yellow = (6*5)/(50*49) rounded to 30/2500
Pink = (4*3)/(49*50) rounded to 12/2500

Then I summed these individual probabilities to get 842/2500
This is ~1/3 as 800*3 = 2400
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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17 Mar 2012, 11:41
1
Its 'C'

(20*19 + 20*19 + 6*5 + 4*3)/50*49 = 1/3
Intern
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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19 Mar 2012, 02:47
why do you use permutation here instead of the 2C50 thing? Im very much confused now Normally I will just go for 2C50 kind of answer whenever I saw this kind of marble selection...
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Posts: 52907
Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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19 Mar 2012, 03:16
2
1
thanhp wrote:
why do you use permutation here instead of the 2C50 thing? Im very much confused now Normally I will just go for 2C50 kind of answer whenever I saw this kind of marble selection...

Though the answer obtained with above formula would be the same, traditional approach would be the one you are talking about.

A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?
A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

$$\frac{C^2_{20}+C^2_{20}+C^2_{6}+C^2_{4}}{C^2_{50}}\approx{\frac{1}{3}}$$, where $$C^2_{20}$$ is ways to select 2 same color balls out of 20 red balls, $$C^2_{20}$$ is ways to select 2 same color balls out of 20 blue balls, $$C^2_{6}$$ is ways to select 2 same color balls out of 6 yellow balls, $$C^2_{4}$$ is ways to select 2 same color balls out of 4 pink balls, and $$C^2_{50}$$ is total ways to select 2 balls out of 50 balls,

Hope it's clear.
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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19 Mar 2012, 03:44
Bunuel wrote:
thanhp wrote:
why do you use permutation here instead of the 2C50 thing? Im very much confused now Normally I will just go for 2C50 kind of answer whenever I saw this kind of marble selection...

Though the answer obtained with above formula would be the same, traditional approach would be the one you are talking about.

A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?
A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

$$\frac{C^2_{20}+C^2_{20}+C^2_{6}+C^2_{4}}{C^2_{50}}\approx{\frac{1}{3}}$$, where $$C^2_{20}$$ is ways to select 2 same color balls out of 20 red balls, $$C^2_{20}$$ is ways to select 2 same color balls out of 20 blue balls, $$C^2_{6}$$ is ways to select 2 same color balls out of 6 yellow balls, $$C^2_{4}$$ is ways to select 2 same color balls out of 4 pink balls, and $$C^2_{50}$$ is total ways to select 2 balls out of 50 balls,

Hope it's clear.

Seriously... Bunuel! You are a real idol for me in all these math things! I have read almost all your posts and replies in Gmatclub today to work on my terrible combination and probability knowledge, and although I still have a long way to an adequate score, your work here means really a lot to me! And thank you again for such a prompt reply! Have a cheerful day!
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Posts: 21
Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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04 Apr 2012, 06:47
Hello Bunuel,

I've a doubt here. Since the q explicitly states w/o replacement shouldn't it be ex. 20 red balls so, 2C20*1C19/50C2 for red again repeat the same for Blue, 2C20*1C19/50C2 +2C20*1C19/50C2 + 2C6*1c5/50c2 + 2C4*1C4/50C2
RED bLUE yELLOW PINK

i HOPE U GOT WAT I'M TRYING TO SAY.

tHNX.
Manager
Joined: 15 Apr 2016
Posts: 71
A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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10 Nov 2016, 05:40
Bunuel wrote:
thanhp wrote:
why do you use permutation here instead of the 2C50 thing? Im very much confused now Normally I will just go for 2C50 kind of answer whenever I saw this kind of marble selection...

Though the answer obtained with above formula would be the same, traditional approach would be the one you are talking about.

A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?
A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

$$\frac{C^2_{20}+C^2_{20}+C^2_{6}+C^2_{4}}{C^2_{50}}\approx{\frac{1}{3}}$$, where $$C^2_{20}$$ is ways to select 2 same color balls out of 20 red balls, $$C^2_{20}$$ is ways to select 2 same color balls out of 20 blue balls, $$C^2_{6}$$ is ways to select 2 same color balls out of 6 yellow balls, $$C^2_{4}$$ is ways to select 2 same color balls out of 4 pink balls, and $$C^2_{50}$$ is total ways to select 2 balls out of 50 balls,

Hope it's clear.

Hi Bunuel,
Just a query..
here in above we have taken RR+BB+YY+PP

should i divide by 4! as i can select either R as first or B as first or Y or P
like BB+YY+PP+RR OR YY+PP+RR+BB etc
am totally confused with this concept.
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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10 Nov 2016, 06:29
1
1
Shrivathsan wrote:
Bunuel wrote:
thanhp wrote:
why do you use permutation here instead of the 2C50 thing? Im very much confused now Normally I will just go for 2C50 kind of answer whenever I saw this kind of marble selection...

Though the answer obtained with above formula would be the same, traditional approach would be the one you are talking about.

A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?
A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

$$\frac{C^2_{20}+C^2_{20}+C^2_{6}+C^2_{4}}{C^2_{50}}\approx{\frac{1}{3}}$$, where $$C^2_{20}$$ is ways to select 2 same color balls out of 20 red balls, $$C^2_{20}$$ is ways to select 2 same color balls out of 20 blue balls, $$C^2_{6}$$ is ways to select 2 same color balls out of 6 yellow balls, $$C^2_{4}$$ is ways to select 2 same color balls out of 4 pink balls, and $$C^2_{50}$$ is total ways to select 2 balls out of 50 balls,

Hope it's clear.

Hi Bunuel,
Just a query..
here in above we have taken RR+BB+YY+PP

should i divide by 4! as i can select either R as first or B as first or Y or P
like BB+YY+PP+RR OR YY+PP+RR+BB etc
am totally confused with this concept.

Dividing by factorial "unarranges", multiplying by factorial accounts for different arrangements. You need neither of these here. We need the cases when we have 2 red, OR 2 blue, OR 2 yellow OR 4 pink balls. We don't have one case of RRBBYYPP which can occur in different ways.

Theory on probability problems

Data Sufficiency Questions on Probability
Problem Solving Questions on Probability

Tough Probability Questions
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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11 Nov 2016, 07:26
tom09b wrote:
A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?

A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

We are given that a jar contains 20 red, 20 blue, 6 yellow and 4 pink balls, so a total of 50 balls. We need to determine the probability that when two balls are selected, they are of the same color. Thus,

P(2 balls of the same color) = P(2 reds) + P(2 blues) + P(2 yellows) + P(2 pinks)

We can use combinations to determine our answers.

Let’s start with the number of ways to select 2 red balls:

20(C)2 = (20 x 19)/2 = 10 x 19 = 190

Since there are 20 blue balls, there are also 190 ways to select 2 blue balls.

Next we can determine the number of ways to select 2 yellow balls:

6(C)2 = (6 x 5)/2 = 3 x 5 = 15

Finally we can determine the number of ways to select 2 pinks balls:

4(C)2 = (4 x 3)/2 =2 x 3 = 6

Thus, the number of ways to select two balls of the same color is:

190 + 190 + 15 + 6 = 401

Now we need to determine the total number of ways to select two balls:

50(C)2 = (50 x 49)/2 = 25 x 49 = 1,125

Thus, P(2 reds) + P(2 blues) + P(2 yellows) + P(2 pinks) = 401/1,125

To more easily reduce our fraction, we can round 401 down to 400 and we have:

400/1,125 = 16/45, which is about 1/3.

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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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23 Nov 2016, 06:35
Bunuel

I tried to do the same as you, (2C20+2C20+2C6+2C4)/C250≈1/3 , however i did the following: (2C20 * 20/50 +2C20 * 20/50 +2C6 * 6/50 +2C4 * 4/50)/C250. Isn't it necessary to include the chances that you take a color?
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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27 Jan 2017, 10:33
Bunuel

What is the difference between with and without replacement? What is the difference between your solution and the solution given by Gavan?

$$\frac{20C2+20C2+6C2+4C2}{50C2}≈\frac{1}{3}$$

and

$$\frac{(20*19 + 20*19 + 6*5 + 4*3)}{50*49} = \frac{1}{3}$$
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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27 Jan 2017, 11:02
Using the 'Slots and Letters' Method, we see that there are 4 ways to get what we want.

RR
BB
YY
PP

Estimating (since the question asks 'approximately' and the answer choices are far apart) the individual probabilities of these sequences, we have:

YY ---> (6/50)(5/49)= (3/25)(about 1/10) or about 3/250, or a very small fraction, not much more that 1/100, negligible here since the answers are so far apart
PP --> If the probability of getting two yellows is very small, the probability of getting two pinks (the least common color) is even smaller, so we can ignore this is well in getting our estimation.

So the answer will be pretty close to the prob of either getting two reds in a row OR two blues in a row. And that will be close to 4/25+4/25 = 8/25 = 32/100.

That's pretty close to answer C, and very far (much further than the errors we introduced could account for) from the other answers.
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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17 Jun 2017, 12:56
Bunuel wrote:
thanhp wrote:
why do you use permutation here instead of the 2C50 thing? Im very much confused now Normally I will just go for 2C50 kind of answer whenever I saw this kind of marble selection...

Though the answer obtained with above formula would be the same, traditional approach would be the one you are talking about.

A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?
A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

$$\frac{C^2_{20}+C^2_{20}+C^2_{6}+C^2_{4}}{C^2_{50}}\approx{\frac{1}{3}}$$, where $$C^2_{20}$$ is ways to select 2 same color balls out of 20 red balls, $$C^2_{20}$$ is ways to select 2 same color balls out of 20 blue balls, $$C^2_{6}$$ is ways to select 2 same color balls out of 6 yellow balls, $$C^2_{4}$$ is ways to select 2 same color balls out of 4 pink balls, and $$C^2_{50}$$ is total ways to select 2 balls out of 50 balls,

Hope it's clear.

What if the balls of the same color are all similar?

I calculated it and got 2/5 as my answer. Is it correct?
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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17 Jun 2017, 13:01
ShashankDave wrote:
Bunuel wrote:
thanhp wrote:
why do you use permutation here instead of the 2C50 thing? Im very much confused now Normally I will just go for 2C50 kind of answer whenever I saw this kind of marble selection...

Though the answer obtained with above formula would be the same, traditional approach would be the one you are talking about.

A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?
A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

$$\frac{C^2_{20}+C^2_{20}+C^2_{6}+C^2_{4}}{C^2_{50}}\approx{\frac{1}{3}}$$, where $$C^2_{20}$$ is ways to select 2 same color balls out of 20 red balls, $$C^2_{20}$$ is ways to select 2 same color balls out of 20 blue balls, $$C^2_{6}$$ is ways to select 2 same color balls out of 6 yellow balls, $$C^2_{4}$$ is ways to select 2 same color balls out of 4 pink balls, and $$C^2_{50}$$ is total ways to select 2 balls out of 50 balls,

Hope it's clear.

What if the balls of the same color are all similar?

I calculated it and got 2/5 as my answer. Is it correct?

It is implied that all balls of the same color are similar. How else?
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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17 Jun 2017, 22:29
tom09b wrote:
A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls. If two balls are to be selected at random and without replacement, approximately what is the probability both of them to be the same colour?

A. 1/25
B. 1/15
C. 1/3
D. 1/2
E. 1

u may select either red or blue or yellow or pink..

hence the result will be (20*19)/50*49 for red ball
similarly for green ball (20*19)/50*49
6*5/50*49 for yellow ball
4*3/50*49 for pink ball

add upp all the four possibilities..
(20*19 + 20*19 + 6*5 + 4*3) / 50*49 ,,approx = 1/3

ans C
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls  [#permalink]

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17 Jan 2019, 17:06
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Re: A jar contains 20 red, 20 blue, 6 yellow and 4 pink balls   [#permalink] 17 Jan 2019, 17:06
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