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A box has some red and some green balls in it. There are no balls of

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A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 30 Sep 2018, 17:33
1
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A
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D
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Difficulty:

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Question Stats:

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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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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 30 Sep 2018, 18:07
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.
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A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 10 Nov 2018, 06:21
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
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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 21 Nov 2018, 10:17
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?
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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 21 Nov 2018, 19:41
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.
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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 17 Jan 2019, 13:48
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
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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 30 Mar 2020, 15:29
chetan2u

I hope you're doing. It has been a while since I tagged you :)

Can you help me understand this better? Even though I got it right in the test I am not confident of my reasoning.
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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 30 Mar 2020, 15:30
LordStark 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%.


Official Explanation

Solution: (C)

Statement (1): If probability of picking a heavy green ball is 20%, that means that 20% of all the balls in the box are heavy and green. But it doesn’t tell us how many balls are heavy and red so statement (1) alone is not sufficient.

Statement (2): If probability of picking a light ball is 40%, it means that 40% of all the balls in the box are light, which means that 60% are heavy. But we do not know how many of them are red, hence statement (2) alone is not sufficient. Both statements together tell us that 60% of all the balls are heavy and 20% of all are heavy and green. This means that 40% of all the balls must be heavy and red. Hence the probability of picking a heavy red ball must be 40%. Both statements together are sufficient to answer the question.
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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 04 Apr 2020, 19:48
1
TheNightKing wrote:
chetan2u

I hope you're doing. It has been a while since I tagged you :)

Can you help me understand this better? Even though I got it right in the test I am not confident of my reasoning.



Hi TheNightKing,

Questions where you have two items and there are two types of each, best is to go for 2 by 2 matrix.

Here two items are Red and Green and each are of two types - light and heavy.

...............R.............G....Total
...L..........a..............b...a+b
...H.........c...............d....c+d
Total....a+c.............b+d.....100

We have to find the value of c

Statement I: d=20...insufficient
Statement II : a+b=40..insufficient

Combined.
Add the two statements —> d+a+b=20+40=60
And a+b+c+d=100, so c=40%
Suff

C
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Re: A box has some red and some green balls in it. There are no balls of  [#permalink]

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New post 04 Apr 2020, 20:24
Thank you for the answers!
YOu really helped me to understand this task.
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Re: A box has some red and some green balls in it. There are no balls of   [#permalink] 04 Apr 2020, 20:24

A box has some red and some green balls in it. There are no balls of

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