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In his pocket, a boy has 3 red marbles, 4 blue marbles, and
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 Updated on: 02 Nov 2012, 13:11
Question Stats:
68% (00:43) correct 32% (00:47) wrong based on 374 sessions
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In his pocket, a boy has 3 red marbles, 4 blue marbles, and 4 green marbles. How many will he have to take out of his pocket to ensure that he has taken out at least one of each color? A. 3 B. 7 C. 8 D. 9 E. 11
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Originally posted by timmaxwell8 on 12 Oct 2009, 11:40.
Last edited by Bunuel on 02 Nov 2012, 13:11, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Probability question
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12 Oct 2009, 12:05
timmaxwell8 wrote: 24. In his pocket, a boy has 3 red marbles, 4 blue marbles, and 4 green marbles. How many will he have to take out of his pocket to ensure that he has taken out at least one of each color?
A. 3
B. 7
C. 8
D. 9
E. 11 The worst scenario would be that he has taken 4 blue and 4 green, total of 8 marbles, and still doesn't have 3 distinct colors. But the next draw (9th) will surely be the third color red as there is no other color marble left in pocket. Answer D (9)
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Re: Probability question
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12 Oct 2009, 20:16
I think, the only way to figure out this problem is just to use logic, just as Bunuel did.
I do not see any other solution for that.
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Re: Probability question
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14 Oct 2009, 03:52
I agree...only logic will help
take out 4G + 4B + 1R = 9 marbles
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Re: Probability question
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06 Jun 2010, 17:39
What about scenario of choosing 3 red and 4 blue and still seeing distinct color.
In other words why can't we have 8 marbles drawn and still see three distinct colors? 3red+4 blue=7 marbles and on 8th draw should be green since remaining should be all green marbles?
Can someone please explain why?
Thanks in advance.
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Re: Probability question
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06 Jun 2010, 19:02
srichaks wrote: What about scenario of choosing 3 red and 4 blue and still seeing distinct color.
In other words why can't we have 8 marbles drawn and still see three distinct colors? 3red+4 blue=7 marbles and on 8th draw should be green since remaining should be all green marbles?
Can someone please explain why?
Thanks in advance. The word "ensure" in the question basically is asking in the worst case scenario. So yes technically, the scenario outlined above is true as well, it is not the worst possible outcome. In order to guarantee the outcome of one of each color, you need to take into account all of the balls in the two largest groups which equals 8. Hope that helped.
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Re: In his pocket, a boy has 3 red marbles, 4 blue marbles, and
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26 Sep 2013, 04:01
What is the difference between this question and the one below.Please explain its confusing:
There are 15 black chips and 5 white chips in a jar. What is the least number of chips we should pick to guarantee that we have 2 chips of the same color?
A. 3
B. 5
C. 6
D. 16
E. 19
Worst case scenario would be if the first two chips we pick will be of the different colors. But the next chip must match with either of two, so 3 is the answer.
Answer: A.
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Re: In his pocket, a boy has 3 red marbles, 4 blue marbles, and
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04 Aug 2019, 03:55
timmaxwell8 wrote: In his pocket, a boy has 3 red marbles, 4 blue marbles, and 4 green marbles. How many will he have to take out of his pocket to ensure that he has taken out at least one of each color?
A. 3
B. 7
C. 8
D. 9
E. 11 9 (4+4+1) worst case
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Re: In his pocket, a boy has 3 red marbles, 4 blue marbles, and
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04 Aug 2019, 06:24
In Such cases, Go for the worst case scenario and hence here we can assume that balls drawn first is (Blue or Green) = 4 and then a single ball to make sure that the guy has atleast one ball of each color which is +1.
Hence the answer is 4+4+1 = 9
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Re: In his pocket, a boy has 3 red marbles, 4 blue marbles, and
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04 Aug 2019, 06:24
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