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There are 12 red balls, 10 blue balls, 15 red balls and 9 [#permalink]

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12 Jun 2012, 23:08

5

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A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

60% (01:55) correct
40% (01:06) wrong based on 109 sessions

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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?

There are 12 red balls, 10 blue balls, 15 red 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. 23 C. 43 D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) 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.

Re: There are 12 red balls, 10 blue balls, 15 red balls and 9 [#permalink]

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13 Jun 2012, 03:15

1

This post received KUDOS

Bunuel wrote:

cyberjadugar wrote:

There are 12 red balls, 10 blue balls, 15 red 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. 23 C. 43 D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) 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.

There are 12 red balls, 10 blue balls, 15 red 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. 23 C. 43 D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) 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.

Re: There are 12 red balls, 10 blue balls, 15 red balls and 9 [#permalink]

Show Tags

14 Jun 2012, 23:15

Bunuel wrote:

cyberjadugar wrote:

There are 12 red balls, 10 blue balls, 15 red 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. 23 C. 43 D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) 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: C.

Sorry, maybe i am missing something, but i cannot understand how we come up to this. So the question is how many balls we need to pick min to make sure that we have on hand at least one colour of each ball! So if we pick all the reds, blues and greens then we have 37 total, but this is not enough since we don't have black, that means if we pick 38, no matter in what way, we will have on hand at least one colour of each ball. So the min number of balls is 38, Could you please tell me where i went wrong in my logic? (most probably i misunderstand the question)
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: There are 12 red balls, 10 blue balls, 15 red balls and 9 [#permalink]

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15 Jun 2012, 00:31

1

This post received KUDOS

ziko wrote:

Sorry, maybe i am missing something, but i cannot understand how we come up to this. So the question is how many balls we need to pick min to make sure that we have on hand at least one colour of each ball! So if we pick all the reds, blues and greens then we have 37 total, but this is not enough since we don't have black, that means if we pick 38, no matter in what way, we will have on hand at least one colour of each ball. So the min number of balls is 38, Could you please tell me where i went wrong in my logic? (most probably i misunderstand the question)

Hi Ziko,

Only flaw in the reasoning is that you are choosing the balls you are picking. If you are lucky, then in 9 chances you may get 9 black balls, but it will not happen every time.

Thus to be sure, considering the worst case scenario, after picking 11 red balls, 9 blue balls, 14 green balls and 8 black balls (total = 42 balls) the next ball you pick would be of any one of these colors. You will definitely have balls of all colors in your \(43^{rd}\)draw.

There are 12 red balls, 10 blue balls, 15 red 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. 23 C. 43 D. 40

Poor quality question. I guess there should be different colors for 12 and 15 balls. Also where is option E?

The worst case scenario would be if we pick 11 red balls, 9 blue balls, 14 (?) 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: C.

Sorry, maybe i am missing something, but i cannot understand how we come up to this. So the question is how many balls we need to pick min to make sure that we have on hand at least one colour of each ball! So if we pick all the reds, blues and greens then we have 37 total, but this is not enough since we don't have black, that means if we pick 38, no matter in what way, we will have on hand at least one colour of each ball. So the min number of balls is 38, Could you please tell me where i went wrong in my logic? (most probably i misunderstand the question)

The point here is that we don't want to have at least one ball of each color, we need ALL balls of some color, so all red balls, or all blue balls, and so on.

Anyway not a good question because of the wording, so I wouldn't worry about it at all.

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