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# Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2

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Director
Joined: 27 May 2008
Posts: 505
Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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04 Aug 2008, 10:36
5
13
00:00

Difficulty:

65% (hard)

Question Stats:

64% (03:07) correct 36% (03:13) wrong based on 122 sessions

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Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2 jars of red paint, and 3 jars of brown paint. Kurt pours the contents of 3 jars of paint into a new container to make a new color, which he will name according to the following conditions:

The paint will be named "Brun Y" if it contains 2 jars of brown paint and no yellow.
The paint will be named "Brun X" if the paint contains 3 jars of brown paint.
The paint will be named "Jaune X" if the paint contains at least 2 jars of yellow.
The paint will be named "Jaune Y" if the paint contains exactly 1 jar of yellow.

What is the probability that the new color will be one of the "Jaune" colors?

A. 5/42
B. 37/42
C. 1/21
D. 4/9
E. 5/9
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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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04 Aug 2008, 11:30
1
durgesh79 wrote:
Kurt, a painter, has 9 jars of paint:
4 are yellow
2 are red
rest are brown
Kurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be Jaune

a) 5/42
b) 37/42
c) 1/21
d) 4/9
e) 5/9

I get B.

Jaune Y = (4 choose 1)*(5 choose 2) = 4*10 = 40
Jaune X = (4 choose 2)*(5 choose 1) + (4 choose 3) = 6*5 + 4 = 34
Total combinations = 9 choose 3 = 84

Probability of Jaune = (40 + 34)/84 = 37/42
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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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02 Feb 2011, 07:32
1
One comment of that:
In the question it states: he will use the collors to make a new color.

But when you mix 3 yellow colors, can that be counted as a new color?
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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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02 Feb 2011, 08:15
3
1
craky wrote:
One comment of that:
In the question it states: he will use the collors to make a new color.

But when you mix 3 yellow colors, can that be counted as a new color?

Well, as OA is B then apparently - yes we can consider the mixture of 3 yellow jars as a new color. Also we are told that we get Jaune X if the paint contains at least 2 jars of yellow, so 3 jars of yellow paint also give Juan X. Though I agree with you that the question is a bit ambiguous (and not only about the issue discussed).

Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2 jars of red paint, and 3 jars of brown paint. Kurt pours the contents of 3 jars of paint into a new container to make a new color, which he will name according to the following conditions:

The paint will be named "Brun Y" if it contains 2 jars of brown paint and no yellow.
The paint will be named "Brun X" if the paint contains 3 jars of brown paint.
The paint will be named "Jaune X" if the paint contains at least 2 jars of yellow.
The paint will be named "Jaune Y" if the paint contains exactly 1 jar of yellow.

What is the probability that the new color will be one of the "Jaune" colors?

A. 5/42
B. 37/42
C. 1/21
D. 4/9
E. 5/9

Basically we are told that we get Jaune in case a container has 1 jar of yellow (Jaune Y), 2 jars of yellow (Jaune X), or all 3 jars of yellow (Jaune X).

So we can calculate the probability that the container does not have yellow paint at all and subtract it from 1: $$P(yellow>0)=1-P(zero \ yellow)=1-\frac{C^3_5}{C^3_9}=\frac{37}{42}$$, where $$C^3_5$$ is the ways we can choose ANY 3 jars out of 2 red+3 brown=5 jars (so not to have yellow) and $$C^3_9$$ is total ways to pick 3 jars out of 9.

Answer: B.
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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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02 Feb 2011, 10:40
B. using the long method of (4c1 * 5c2 + 4c2 * 5c1 + 4c3)/9c3 = 37/42
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Joined: 15 Mar 2012
Posts: 40
Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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25 Jun 2014, 04:04
Experts, could you please explain what's wrong with the following approach:

4 yellow, 2 red, 3 brown jars are available

p(J) = p(JX) + p (JY)
= (yellow * yellow * any) + (yellow * any other * any other)
= (4/9 * 3/8 * 7/7) + (4/9 * 5/8 * 4/7)
= 41/126 --> gives me the wrong anwers
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Joined: 02 Sep 2009
Posts: 55804
Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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25 Jun 2014, 07:48
divineacclivity wrote:
Experts, could you please explain what's wrong with the following approach:

4 yellow, 2 red, 3 brown jars are available

p(J) = p(JX) + p (JY)
= (yellow * yellow * any) + (yellow * any other * any other)
= (4/9 * 3/8 * 7/7) + (4/9 * 5/8 * 4/7)
= 41/126 --> gives me the wrong anwers

3 mistakes there:

1. You forgot to consider the (yellow, yellow, yellow) case
2. In (yellow * yellow * any) case the last fraction should be 5/7, not 7/7. You want to choose any non-yellow paint from 7 jars left.
3. You need to apply factorial correction: YYA can occur in 3 way YYA, YAY, and AYY. Similarly, YAA can occur in 3 ways: YAA, AYA, and YYA.

So, the correct formula should be: $$(\frac{4}{9} * \frac{3}{8} * \frac{5}{7})*\frac{3!}{2!} + (\frac{4}{9} * \frac{5}{8} * \frac{4}{7})*\frac{3!}{2!}+\frac{4}{9}*\frac{3}{8}*\frac{2}{7}=\frac{37}{42}$$.

Hope it's clear.
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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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26 Jun 2014, 01:49
Bunuel wrote:
divineacclivity wrote:
Experts, could you please explain what's wrong with the following approach:

4 yellow, 2 red, 3 brown jars are available

p(J) = p(JX) + p (JY)
= (yellow * yellow * any) + (yellow * any other * any other)
= (4/9 * 3/8 * 7/7) + (4/9 * 5/8 * 4/7)
= 41/126 --> gives me the wrong anwers

3 mistakes there:

1. You forgot to consider the (yellow, yellow, yellow) case
2. In (yellow * yellow * any) case the last fraction should be 5/7, not 7/7. You want to choose any non-yellow paint from 7 jars left.
3. You need to apply factorial correction: YYA can occur in 3 way YYA, YAY, and AYY. Similarly, YAA can occur in 3 ways: YAA, AYA, and YYA.

So, the correct formula should be: $$(\frac{4}{9} * \frac{3}{8} * \frac{5}{7})*\frac{3!}{2!} + (\frac{4}{9} * \frac{5}{8} * \frac{4}{7})*\frac{3!}{2!}+\frac{4}{9}*\frac{3}{8}*\frac{2}{7}=\frac{37}{42}$$.

Hope it's clear.

Thank you for your reply, Dear Bunuel!

I took 7/7 instead of 5/7 & 2/7 separately thinking this would cover YYY as remaining 7 include 2 of yellow as well apart from 5 of B & R; thas why I mentioned "any" and not "any other".

Nevertheless, I got you point.
Y = yellow
A = Any other
What you're saying is that the following is how it should be:

{(4x3/2!)x5}/(9x8x7/3!) + (4x3x2/3!)/(9x8x7/3!) + {(4x(5x4)/2!}/(9x8x7/3!)
2!, 3! are to exclude the order of selection or as you called it "factorial correction"

Now, what I'm wondering how I was able to do all such probability questions right without consindering factorial correction (e.g. 3!/2!). I did those right probably because factorial corection (exclusion of count of the oder of selection) was automatically taken care of, as for both numerator and denominator, it must have been the same e.g. 3!/3! or 2!/2! OR I must've calculated the numerator (count of favourable events) and denominator (count of total events) separately, in which I always apply factorial correction for combinations.

e.g. If I had to calculate probability of selection of 3 y from 3 y & 2 r, I just did 3/5 * 2/4 * 1/3. It was always right because factorial correction was the same for both numerator and denominator 3!/3! i.e. (3*2*3/3!)/(5*4*3/3!)
Am I right?
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Joined: 02 Sep 2009
Posts: 55804
Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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26 Jun 2014, 01:57
divineacclivity wrote:
Bunuel wrote:
divineacclivity wrote:
Experts, could you please explain what's wrong with the following approach:

4 yellow, 2 red, 3 brown jars are available

p(J) = p(JX) + p (JY)
= (yellow * yellow * any) + (yellow * any other * any other)
= (4/9 * 3/8 * 7/7) + (4/9 * 5/8 * 4/7)
= 41/126 --> gives me the wrong anwers

3 mistakes there:

1. You forgot to consider the (yellow, yellow, yellow) case
2. In (yellow * yellow * any) case the last fraction should be 5/7, not 7/7. You want to choose any non-yellow paint from 7 jars left.
3. You need to apply factorial correction: YYA can occur in 3 way YYA, YAY, and AYY. Similarly, YAA can occur in 3 ways: YAA, AYA, and YYA.

So, the correct formula should be: $$(\frac{4}{9} * \frac{3}{8} * \frac{5}{7})*\frac{3!}{2!} + (\frac{4}{9} * \frac{5}{8} * \frac{4}{7})*\frac{3!}{2!}+\frac{4}{9}*\frac{3}{8}*\frac{2}{7}=\frac{37}{42}$$.

Hope it's clear.

If I had to calculate probability of selection of 3 y from 3 y & 2 r, I just did 3/5 * 2/4 * 1/3. It was always right because factorial correction was the same for both numerator and denominator 3!/3! i.e. (3*2*3/3!)/(5*4*3/3!)
Am I right?

In this case yyy can occur only in one way, so no need for factorial correction.

If it were what is the probability of selecting 2 y's and 1 r out of {yyyrr}, then the probability would be P(yyr) = 3/5*2/4*2/3*3!/2!, because yyr can occur in 3!/2!=3 ways: yyr, yry, ryy.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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26 Jun 2014, 02:53
Bunuel wrote:
In this case yyy can occur only in one way, so no need for factorial correction.

If it were what is the probability of selecting 2 y's and 1 r out of {yyyrr}, then the probability would be P(yyr) = 3/5*2/4*2/3*3!/2!, because yyr can occur in 3!/2!=3 ways: yyr, yry, ryy.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

Thank you very much. That helps a lot.
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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2  [#permalink]

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29 Apr 2019, 07:27
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Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yellow paint, 2   [#permalink] 29 Apr 2019, 07:27
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