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Tuck School Moderator P
Status: Valar Dohaeris
Joined: 31 Aug 2016
Posts: 301
GMAT 1: 700 Q49 V37 A box has some red and some green balls in it. There are no balls of  [#permalink]

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1
3 00:00

Difficulty:   15% (low)

Question Stats: 79% (01:22) correct 21% (01:09) wrong based on 158 sessions

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A box has some red and some green balls in it. There are no balls of any other color. Some of these balls are heavy and the rest are all light. What is the probability that a heavy red ball is picked out of the box?

1. The probability of picking a heavy green ball is 20%
2. The probability of picking a light ball is 40%.

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Manager  S
Joined: 21 Nov 2016
Posts: 70
GMAT 1: 640 Q47 V31 Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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3
For the sake of convenience lets denote,
Light green = lg
Heavy green = hg
Light red = lr
Heavy red = hr
lg+hg+lr+hr = 100 % (say 100 balls)
Implies, lg+hg+lr+hr = 100

From statement 1 we have that the probability of picking up a hg is 20% . That is there are 20 balls which are heavy green out of those 100 balls.
Implies , hg+ lg+ lr+ hr = 100,
Implies, 20+ lg+ lr+ hr = 100,
Implies, lg + lr + hr = 80. But from this equation we do not know the value of heavy red ie hr.
So, insufficient.

Statement 2 says probability of picking a light ball is 40% that is there are 40 balls which are light out of those 100 balls.
So, hg + hr + lr + lg = 100
We have from statement 2, lg + lr = 40
So, hg + hr +40 = 100
Hg+ Hr = 60. Still we cannot find Hr.
So, Insufficient
.
From statement 1 we have , Hg = 20, so, substituting in the result of statement 2, we get , 20 + Hr = 60. Implies Hr = 40.
So the correct option is C.
Intern  B
Joined: 18 Mar 2018
Posts: 27
A box has some red and some green balls in it. There are no balls of  [#permalink]

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This can be done by using a comparison table:

As no number has been assigned for the number of balls, pick a smart number of 100 for simplicity

-------R-----G------T
H-----x--------------
L---------------------
T------------------100

Statement 1: You have heavy and green balls, but no information on the number of light balls; NOT SUFFICIENT

-------R-----G------T
H-----x-----20---20+x
L---------------------
T------------------100

Statement 2: You can determine the total number of heavy balls from this but no information from the Green and Red split; NOT SUFFICIENT

-------R-----G-----T
H-----x-----60-x 60
L------------------40
T------------------100

Statement 1 & 2: Combining the 2, you have enough information by simply filling in the table

-------R-----G-----T
H-----x-----20---60
L------------------40
T------------------100

x = 60-20 = 40
Manager  G
Joined: 13 Nov 2018
Posts: 118
Location: India
GMAT 1: 700 Q51 V32 Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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PeepalTree wrote:
A box has some red and some green balls in it. There are no balls of any other color. Some of these balls are heavy and the rest are all light. What is the probability that a heavy red ball is picked out of the box?

1. The probability of picking a heavy green ball is 20%
2. The probability of picking a light ball is 40%.

how c is correct and why not A

Prob(heavy green ball)=0.2

can we write

Prob(heavy red ball)=1-P(heavy green ball)

anyone can please explain simple way?
Intern  B
Joined: 03 Sep 2018
Posts: 1
Location: India
Concentration: General Management
GMAT 1: 640 Q47 V31 GPA: 3.6
WE: Engineering (Energy and Utilities)
Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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gmatdordie

You missed out on including lighter balls

Prob(heavy red ball) + Prob(light red ball) + P(heavy green ball) + Prob(light green ball) = 1

Soln:

P(heavy green) = 1/5
P(light) = 2/5 => P(heavy) = 3/5

P(heavy green) = P(heavy) x P(green) [picking a ball based on weight and picking a ball based on color are mutually exclusive) => P(green) = 1/3 => P(red) = 2/3

P(heavy red) = P(heavy) x P(red) = 3/5 x 2/3 = 2/5

Bunuel can you crosscheck my reasoning. Thanks.
Intern  B
Joined: 26 Dec 2018
Posts: 25
Concentration: Finance, Economics
Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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1. P(HG)=.2
1-.2=.8=P(HR)+P(LR)+P(LG)

No way of determining probabilities of other marble groupings(insufficient).

2. 1-P(L)=.6=P(H)

On its own,insufficient.

1&2(combined).

P(HG)=P(H)*P(G)

.2=.6*P(G)

P(G)=.33

1-.33=.67=P(R)

P(R)*P(H)=P(HR)

.67*.6=.402 (sufficient Re: A box has some red and some green balls in it. There are no balls of   [#permalink] 17 Jan 2019, 14:48
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