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# Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y

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Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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05 Oct 2012, 10:26
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Difficulty:

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Question Stats:

63% (03:25) correct 37% (03:56) wrong based on 108 sessions

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Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are yellow, and the rest are brown. Charlie will combine 3 jars of paint into a new container to make a new colour, which he will name according to the following conditions:

1) C1, if the paint contains 2 jars of brown paint and no blue paint
2) C2, if the paint contains 3 jars of brown paint.
3) J1, if the paint contains at least 2 jars of blue paint
4) J2, if the paint contains exactly 1 jar of blue paint

What is the probability that the new colour will be a shade of J (J1 or J2)?

(A) 75/84
(B) 10/21
(C) 17/42
(D) 11/21
(E) 37/42

For me is a good question, maybe time consuming because when you understand what is asked, the question must be set-up carefully.

Let me know
[Reveal] Spoiler: OA

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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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05 Oct 2012, 13:21
[/spoiler]
carcass wrote:
Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are yellow, and the rest are
brown. Charlie will combine 3 jars of paint into a new container to make a new
colour, which he will name according to the following conditions:
1) C1, if the paint contains 2 jars of brown paint and no blue paint
2) C2, if the paint contains 3 jars of brown paint.
3) J1, if the paint contains at least 2 jars of blue paint
4) J2, if the paint contains exactly 1 jar of blue paint
What is the probability that the new colour will be a shade of J (J1 or J2)?
(A) 75/84
(B) 10/21
(C) 17/42
(D) 11/21
(E) 37/42

For me is a good question, maybe time consuming because when you understand what is asked, the question must be set-up carefully.

Let me know

Probability of J=Probability of J1 + J2

Probability of J1=((4c2*5c1) + (4c3))/(9c4)----------------atleast 2 so it can be 2 blue or 3 blue
solving we get.
=(17/42)

Probability of J2=(4c1*5c2)/(9c4)
solving we get
=(20/42)
Probability of j=(17/42) + (20/42)
=37/42

ans is E

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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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05 Oct 2012, 15:36
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carcass wrote:
Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are yellow, and the rest are
brown. Charlie will combine 3 jars of paint into a new container to make a new
colour, which he will name according to the following conditions:
1) C1, if the paint contains 2 jars of brown paint and no blue paint
2) C2, if the paint contains 3 jars of brown paint.
3) J1, if the paint contains at least 2 jars of blue paint
4) J2, if the paint contains exactly 1 jar of blue paint
What is the probability that the new colour will be a shade of J (J1 or J2)?
(A) 75/84
(B) 10/21
(C) 17/42
(D) 11/21
(E) 37/42

For me is a good question, maybe time consuming because when you understand what is asked, the question must be set-up carefully.

Let me know

If the new color doesn't contain Blue paint, then it isn't of type J. So, it is much easier to calculate the complementary probability.
P(no Blue paint chosen) = (5/9)*(4/8)*(3/7) = 5/42.
Therefore, the required probability is 1 - 5/42 = 37/42.

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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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05 Oct 2012, 16:51
Evajager your approach is really fast

The first one you have to calculate the various combination (formula) with j1 and j2

thanks
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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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10 Oct 2012, 05:29
1
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Using Combinatrix

I start the same way as EvaJager

What is the probability that the new paint has either 1,2, or 3 parts blue in it.

This is equivalent of saying 1-probability(no blue)

First, how many ways can we choose 3 paints from 9 paints 3C9, 9!/(3!6!)=84

Second how many ways can we make a paint with no blue, or choose 3 paints from the remaining 5 paints 3C5 = 5!/(3!2!) = 10

answer is 1 -(10/84) = 74/84.

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Re: Charlie, a painter, has 9 jars of paint…700 Level PS [#permalink]

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12 Apr 2013, 21:12
Solution 1:
Number of ways of making J1 = number of ways of choosing the paints such that there are two blue paints + number of ways of choosing the paints such that there are three blue paints = 4C2 x 5C1 + 4C3 = 34

Number of ways of making J2 = number of ways of choosing the paints such that there is one jar of blue and two jars of any other color = 4C1 x 5C2 = 40

Number of ways of combining three paints to make a color = 9C3 = 84

Number of ways (new color will be a shade of J1 or J2) = n(J1) + n(J2) - n(J1 n J2) = n(J1) + n(J2) as n(J1 n J2) = 0 [either there will be exactly one color of blue or two or more]

Therefore probability (new color will be a shade of J1 or J2) = (40+34)/84 = 74/84 = 37/42
Option E

Solution 2:
Looking at the classifications, we realize straight away that J1 or J2 = number of ways of selecting the paints such that there is at least one blue paint included
Therefore probability = 1 - prob. (no blue paint is included)
= 1 - (5C3/9C3) = 1 - 10/84 = 74/84 = 37/42
Option E
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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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13 Apr 2013, 13:42
Think about it like this: the question is asking for the probability that we have three jars of paint, none of which are blue.

Thus, it is basically an "at least" question, which invariably implies 1 (full probability) - (probability of the inverse occurrence).

That is, if you want the probability that a coin flipped three times will land on Heads at least once, calculate the probability of landing ONLY ON TAILS (=1/8) and subtract it to find the "at least once" situation.

1 - 1/8 = 7/8 chance that it will land on Heads at least once.

So it is simplest to think about it as asking "what is the probability that the mix will have at least one jar of blue paint in it?"

The probability of having no jars of blue paint is (-B)(-B)(-B) = (5/9)(4/8)(3/7), which reduces to 5/42.

That is, 1 - no blue = 1 - (-B)(-B)(-B) = 1 - 5/42 = 37/42. The answer is E.
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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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16 Jul 2014, 14:29
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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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13 Feb 2016, 03:59
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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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15 Feb 2016, 09:48
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Please let me know where am i going wrong,
the probability that it would be J1 or J2 is same as 1-()probability that it is C1 or C2

For C1-3C2(Brown)*2C1(Yellow)=3*2=6
For C2-3C3(all brown)=1

Total=9C3=84
1-7/84=77/84

I am confused
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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y [#permalink]

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25 Apr 2016, 12:50
There are three possibilities of getting J shades.

1. Two Blue and 1 non-blue = 4C2 * 5C1 = 6 * 5 = 30
2. Three blue = 4C3 = 4
3. One Blue and two non-blue = 4C1 * 5C2 = 4 * 10 = 40

Total number of possibilities = 9C3 = 84

(40+4+30)/84 = 74/84.

ANS E.

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Re: Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are y   [#permalink] 25 Apr 2016, 12:50
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