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There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball

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Bunuel
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There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 04:42
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There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 16
C. 23
D. 42
E. 43
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Bunuel
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There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 04:43
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Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 16
C. 23
D. 42
E. 43


The worst-case scenario would be if we pick 11 red balls, 9 blue balls, 14 green balls, and 8 black balls. In this case, we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next ball, the 43rd one we pick, no matter which color it is, will guarantee that we have all the balls of one color.


Answer: E.
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chetan2u
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 06:56
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testcracker wrote:
chetan2u wrote:
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 16
C. 23
D. 42
E. 43



In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+9-4+1=43


hi

do you think 38 can also be an answer, although the choice is not in the answer choice ?

say first we pick all 15 green balls, then all 12 red balls, and then all 10 blue balls, so lastly we end up with only black balls remaining

so 15 + 12 + 10 +1 = 38

so, where am I getting wrong ?

thanks


The problem is with understanding the question, which can become quite perplexed.

The question says the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color.
The catch is ASSURED. I can pick up just 9 all and all 9 can be blacks and that will be the best scenario.

But the word ASSURED means that the probability of one colour has to be 100%. These 9 could actually be combinations of all 4 colours.
This means that I pick up some number blindly and then be able to claim that I have picked up atleast one set of colours, and here we will consider the worst scenario.
That is I have picked up all but 4, and these 4 are one of each colour.
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   01 Feb 2019, 19:01
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Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 16
C. 23
D. 42
E. 43


The worst case scenario is that we take out 14 green balls, 11 red balls, 9 blue balls and 8 black balls. We see that we still don’t have all the balls of one color. However, if we take out one more ball, no matter what color it is, we will have all the balls of one color. Therefore, the total number of balls need to be taken out is 14 + 11 + 9 + 8 + 1 = 43.

Answer: E
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 05:06
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Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 16
C. 23
D. 42
E. 43



In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+9-4+1=43
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There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   Updated on: 30 Jan 2019, 07:03
chetan2u wrote:
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 16
C. 23
D. 42
E. 43



In other words, the worst case is when the last 4 are all of different colour, and the one you pick will result in picking all of one type, so ans = 12+10+15+9-4+1=43


hi

do you think 38 can also be an answer, although the choice is not in the answer choice ?

say first we pick all 15 green balls, then all 12 red balls, and then all 10 blue balls, so lastly we end up with only black balls remaining

now, 15 + 12 + 10 +1 = 38

so, where am I getting wrong ?

thanks

Originally posted by testcracker on 30 Jan 2019, 06:47.
Last edited by testcracker on 30 Jan 2019, 07:03, edited 1 time in total.
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 07:46
Bunuel wrote:
Bunuel wrote:
There are 12 red balls, 10 blue balls, 15 green balls and 9 black balls in a bag. What is the minimum number of balls to be taken out of the bag to assure that you have all the balls of one color?

A. 10
B. 16
C. 23
D. 42
E. 43


The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 green balls and 8 black balls. In this case we'll have 11+9+14+8=42 balls and still won't have all the balls of one color. The next 43rd ball we pick, no matter which color it'll be, will guarantee that we have all the balls of one color.

Answer: E.

Outstanding. Thank you
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 10:02
shouldn't the question say "maximum number of balls" instead of minimum?
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 10:11
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arpitkansal wrote:
shouldn't the question say "maximum number of balls" instead of minimum?



No, it will not be maximum. There has to be some contrast in two things..

I can say what will be the maximum number of balls that can be drawn without getting all balls of any one colour out.
Here the max will become all except one of each left, so 42.

But if we use maximum in present case. Maximum can be 47, as we will hav all of all colour. So here we have minimum such that one entire colour is present..

Contrast .. maximum still all not there.
Minimum such that one is there.

Hope it helps
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 10:55
But Chetan,the minimum could be that I draw 9 black balls simultaneously and then I would have fulfilled the condition of "have all the balls of one color"..I am still confused.
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   30 Jan 2019, 11:17
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arpitkansal wrote:
But Chetan,the minimum could be that I draw 9 black balls simultaneously and then I would have fulfilled the condition of "have all the balls of one color"..I am still confused.


Arpit, the point is of being 100% sure, because there is a word ASSURED.
Can you pick 9 balls without seeing and announce that all are of same colour.. No you cannot
.
You cannot till the time you don't take the worst scenario..

So if you pick 43, 44, 45 or 46, you will be 100% sure that you have atleast one colour completely with you and yes you can claim without hesitation.
MINIMUM comes into play in these 4 numbers, so 43 is the minimum here and our answer
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink] New post   14 Aug 2023, 06:39
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Re: There are 12 red balls, 10 blue balls, 15 green balls and 9 black ball [#permalink]
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