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Kurt, a French painter, has 9 jars of paint: 4 jars of yello [#permalink]
04 Aug 2008, 09:36

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This post was BOOKMARKED

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Difficulty:

45% (medium)

Question Stats:

51% (02:32) correct
49% (02:17) wrong based on 46 sessions

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?

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

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.

Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yello [#permalink]
25 Jun 2014, 06:48

Expert's post

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}.

Kurt, a French painter, has 9 jars of paint: 4 jars of yello [#permalink]
26 Jun 2014, 00:49

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?

Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yello [#permalink]
26 Jun 2014, 00:57

Expert's post

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.

Re: Kurt, a French painter, has 9 jars of paint: 4 jars of yello [#permalink]
26 Jun 2014, 01:53

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.