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Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks

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Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256
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Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256


\(P(0 < blue < 4) = 1 - P(blue \ = \ 0 \ or \ 4) =\)

\(=1 - (\frac{75}{100}*\frac{75}{100}*\frac{75}{100}*\frac{75}{100} + \frac{25}{100}*\frac{25}{100}*\frac{25}{100}*\frac{25}{100}) = \frac{87}{128}\).

Answer: C.
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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 07 Oct 2014, 06:59
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256


1-P(not B)= probability that at least 1 blue disk will be selected.
p(not B) = no blue disk is selected in each draw.
= 1- (75/100)(75/100)(75/100)(75/100)
=1-81/256
=175/256

now this probability also includes the case , when all of the selected disc at each of the four draw will be blue. so we have to subtract this case from this probability to get the final answer.

probability that only blue disc got selected at each draw= (25/100)(25/100)(25/100)(25/100)
=1/256

thus the required probability = (175-1)/256 =174/256
=87/128

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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Dear sharmasneha,
Thanks for posting this question. +1 to you. Here's my take on this

Probability of getting no less than 1 and no more than 3 blue disks = 1-[ Probability of getting 0 blue disks + Probability of getting 4 blue disks]
Prob. of getting 0 blue disks in one bag = 75/100
Prob. of getting 0 blue disks in all bags = (75/100)^4

Prob of getting 4 blue disks in one bad = 25/100
Prob. of getting 4 blue disks in all bags = (25/100)^4

Adding the two = 82/256 = 41/128

1 - 41/128 = 87 / 128
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Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 07 Oct 2014, 22:38
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 andno greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256



I have realized that we should multiply all non scenarios because problem says no less AND no more. If it said OR we could add and get above stated answers. My own speculation it that, because we are saying about undesirable conditions, we should add to get higher probability and subtract it from 1. Could experts explain, please

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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Temurkhon wrote:
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 andno greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256



I have realized that we should multiply all non scenarios because problem says no less AND no more. If it said OR we could add and get above stated answers. My own speculation it that, because we are saying about undesirable conditions, we should add to get higher probability and subtract it from 1. Could experts explain, please


Dear Temurkhon

I understand that what confused you is a bit tricky word - AND - used in the question. Here's my take on it.

what is the probability that the number of blue disks chosen will be no less than 1 andno greater than 3

The above statement means that we are looking for 1, 2 and 3. If we do 1-(P0 * P4) we are entering an impossible territory. It is not possible to have both the conditions at the same time i.e. We can't have P0 and P4 at the same time.

Lets understand it this way :

We have the following probabilities for blue :
P0
P1
P2
P3
P4

and at a time, we can have only one of these, i.e. they are mutually exclusive.

SO, to eliminate the possibility of P0 AND P4, we need to add them and then eliminate the chances of their occurrence.

Hope that helps you.
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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 08 Oct 2014, 00:48
As I see, in this case AND just means the desirable area, i.e. 1 to 3 blue balls, but not two desirable conditions that we should combine.

Thanks

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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Temurkhon wrote:
As I see, in this case AND just means the desirable area, i.e. 1 to 3 blue balls, but not two desirable conditions that we should combine.

Thanks


Dear Temurkhon,

Your understanding is correct. Since you used a very interesting word, AREA, I presume you have set theory in your mind. Just to make things a bit more clear, have a look at the figure.

If you look at the figure, the five circles tell you the area of the possibility of each condition. We will never have a condition in which we can have 2 and 3 blue balls.

Its either 2 blue balls or 3 blue balls.

Thus there would never be an intersection between the circles.

Since there is no intersection between any of the sets, we can't multiply any of the probability, and that's why we call them mutually exclusive.

Hope that helps you.
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Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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I did it the long way

Prob of getting 1 , 2 , or 3 Blue balls.

1 ball: 4C1 (1/4) (3/4)^3
2 ball: 4C2 (1/4)^2 (3/4)^2
3 balls: 4C3 (1/4)^3 (3/4)

Adding all = (108+54+12)/(4 X4 X4X4) = 87/128.

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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experts whats wrong in this

p of one blue + p of 2 blue + p of 3 blue

1/4*3/4*3/4*3/4 +1/4*1/4*3/4*3/4 + 1/4*1/4*1/4*3/4

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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vipulgoel wrote:
experts whats wrong in this

p of one blue + p of 2 blue + p of 3 blue

1/4*3/4*3/4*3/4 +1/4*1/4*3/4*3/4 + 1/4*1/4*1/4*3/4


Hi vipulgoel,
you are correct to some extent..
where you rae going wrong is
1) p of one blue should be multiplied by 4C1 as you can cchoose one bag containing blue in 4C1 ways..
2) similarly p of 2 blue should be multiplied by 4C2 as these two bags can be choosen in 4C2 ways..
3) p of 3 blue by 4C3..
you will get your answer..

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 07 Jun 2016, 21:33
Bunuel wrote:
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256


\(P(0 < blue < 4) = 1 - P(blue \ = \ 0 \ or \ 4) =\)

\(=1 - (\frac{75}{100}*\frac{75}{100}*\frac{75}{100}*\frac{75}{100} + \frac{25}{100}*\frac{25}{100}*\frac{25}{100}*\frac{25}{100}) = \frac{87}{128}\).

Answer: C.



Hi Bunuel,

Since the problem does not explicitly state that we will replace the disks that are taken out of the bags, shouldn't we have accounted for the non-replacement?

I was thinking along the line of this example; we pick 2 blocks from a box holding 10 blocks, 3 of which are red. The probability of getting 2 red blocks then is 3/10*2/9= 1/15. Using this approach on the question of interest, I arrived at a very ugly figure which was definitely not the OA.

Thank you.

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 08 Jun 2016, 01:49
TheLostBear wrote:
Bunuel wrote:
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256


\(P(0 < blue < 4) = 1 - P(blue \ = \ 0 \ or \ 4) =\)

\(=1 - (\frac{75}{100}*\frac{75}{100}*\frac{75}{100}*\frac{75}{100} + \frac{25}{100}*\frac{25}{100}*\frac{25}{100}*\frac{25}{100}) = \frac{87}{128}\).

Answer: C.



Hi Bunuel,

Since the problem does not explicitly state that we will replace the disks that are taken out of the bags, shouldn't we have accounted for the non-replacement?

I was thinking along the line of this example; we pick 2 blocks from a box holding 10 blocks, 3 of which are red. The probability of getting 2 red blocks then is 3/10*2/9= 1/15. Using this approach on the question of interest, I arrived at a very ugly figure which was definitely not the OA.

Thank you.


If there is a replacement it'll be explicitly mentioned. Also, we are choosing only 1 disk from each bag, so replacement does not play part in this scenario at all.
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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 08 Jun 2016, 11:38
Bunuel wrote:
TheLostBear wrote:
Bunuel wrote:

Hi Bunuel,

Since the problem does not explicitly state that we will replace the disks that are taken out of the bags, shouldn't we have accounted for the non-replacement?

I was thinking along the line of this example; we pick 2 blocks from a box holding 10 blocks, 3 of which are red. The probability of getting 2 red blocks then is 3/10*2/9= 1/15. Using this approach on the question of interest, I arrived at a very ugly figure which was definitely not the OA.

Thank you.


If there is a replacement it'll be explicitly mentioned. Also, we are choosing only 1 disk from each bag, so replacement does not play part in this scenario at all.

Got it. Can I safely assume replacement while doing similar problems on the GMAT then, unless stated otherwise?

Also, maybe I'm reading too much into the question, but since 25 blue disks can't be evenly distributed to four bags, wouldn't one bag have higher probability of drawing a blue disk than the others? In that case, even though replacement doesn't play a part, wouldn't the answer be different?

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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Bunuel wrote:
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256


\(P(0 < blue < 4) = 1 - P(blue \ = \ 0 \ or \ 4) =\)

\(=1 - (\frac{75}{100}*\frac{75}{100}*\frac{75}{100}*\frac{75}{100} + \frac{25}{100}*\frac{25}{100}*\frac{25}{100}*\frac{25}{100}) = \frac{87}{128}\).

Answer: C.


I did this one like this.
We know that each of the bags contain equal number of all colors i.e. selection of ball of every color is equally likely. Thus, this question can also be done with kind of a binomial method.

we have four places to fill
__ __ __ __

Each of the places can be B,G,O or Y.

Total number of ways
\(4^4 = 256\)

Number of ways in which none are B

\(3^4 = 81\)

Number of ways in which all are B

\(1\)

we have to subtract this total from all the possibilities

\(256-82 = 174\)

Answer

\(\frac{178}{256}\)

==\(\frac{87}{128}\)

(C)

Bunuel, is the reason why I have applied this method here correct?
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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 07 Nov 2017, 04:42
rukna wrote:
I did it the long way

Prob of getting 1 , 2 , or 3 Blue balls.

1 ball: 4C1 (1/4) (3/4)^3
2 ball: 4C2 (1/4)^2 (3/4)^2
3 balls: 4C3 (1/4)^3 (3/4)

Adding all = (108+54+12)/(4 X4 X4X4) = 87/128.


I dont understand from where 4C1, 4C2 and 4C3 come from? They are not used when applying 1-P(0) or P(4). Then why here?

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Re: Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks [#permalink]

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New post 08 Nov 2017, 16:40
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 and no greater than 3?

(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256


We should first note that the condition “the number of blue disks chosen is no less than 1 and no greater than 3” is equivalent to the condition “the number of blue disks is neither 0 nor 4.”

We can use the following equation:

P(the number of blue disks chosen will be no less than 1 and no greater than 3) = 1 - P(selecting 0 blue disks) - P(selecting 4 blue disks)

Let’s first determine P(selecting 0 blue disks):

75/100 x 75/100 x 75/100 x 75/100 = (3/4)^4 = 81/256

Next let’s determine P(selecting 4 blue disks):

25/100 x 25/100 x 25/100 x 25/100 = (1/4)^4 = 1/256

Thus:

P(the number of blue disks chosen will be no less than 1 and no greater than 3) is:

1 - 81/256 - 1/256 = 1 - 82/256 = 174/256 = 87/128

Answer: C
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