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# The last one probe for today: There are 8 balls in a box;

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SVP
Joined: 03 Feb 2003
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The last one probe for today: There are 8 balls in a box; [#permalink]  27 Jun 2003, 02:11
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The last one probe for today:

There are 8 balls in a box; among them white, red, blue, and black ones. The rest four are colorless. Four balls are taken at random without repetition. What is the probability of having all the colored balls?

Last edited by stolyar on 16 Jul 2003, 04:20, edited 1 time in total.
SVP
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Kudos [?]: 75 [0], given: 0

I assume that there are four colorless balls in the box besides the four having any color. Thus, the question is an example of hypergeometric distribution.

P=1C1*1C1*1C1*1C1/8C4=1/70
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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Kudos [?]: 51 [0], given: 0

stolyar wrote:
I assume that there are four colorless balls in the box besides the four having any color.

How can you make this assumption based on the wording of the question? No where does it state that there is only one ball of each of the four colors.

The probability will certainly change according to the numbers of each type of ball in the bag. Unless, of course, you are asserting that the average probability, after considering the plethora of possible combinations are colored balls given that at least one of each color is in the bag, will reduce to that of having 4 colorless balls. But that cannot be the case, because any time you repeat the number of balls, you increase the number of ways you can pick out 4 different colored balls. So clearly, 1/70 is the smallest possible probability and hence cannot possible be the average.

It is *possible* to solve this problem given no information other than that this is at least one ball of each color and that the other balls each have an equal change of being one of the 5 colors, the four or a neutral one. But it will be very difficult and timeconsuming to do so because it involve summing up a large number of weighted probabilities based on the various possible distributions in the bag.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

I agree with your criticism. Forgive me.
I can only say that that day I was pretty tired inventing questions, and that this one was the 20th or 21st one.

Eliminate this question?
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Followers: 13

Kudos [?]: 51 [0], given: 0

stolyar wrote:
I agree with your criticism. Forgive me.
I can only say that that day I was pretty tired inventing questions, and that this one was the 20th or 21st one.

Eliminate this question?

Just rewrite the questions so that the assumptions are clear. BTW, i KNOW firsthand how hard it is to write good questions so kudos to you for even trying.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Eternal Intern
Joined: 07 Jun 2003
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Kudos [?]: 43 [0], given: 0

Hypogeometric Probability [#permalink]  16 Jul 2003, 09:14
What is that?

Also four balls are taken without replacement, if we said four at one chunk, it would be 1/2.

Please respond to message I left you on PS:Shoes,
Eternal Intern
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Stolyar-the master [#permalink]  17 Jul 2003, 06:44
Four balls are taken at random without repetition: You wrote Stolyar.

How about four balls taken without replacement.

C ( 1 ball, 1 colorless) * ..... C( 1 ball, 1 colorless)
One colorless ball cannot be replaced, but there are three others bringing the total to four !-

VT
Director
Joined: 28 Oct 2003
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Kudos [?]: 10 [0], given: 0

4/8 * 3/7 * 2/6 * 1/5

24/(48*35)

1/70

Which is what anandk has, but expressed differently.

anand-- I just took the probability of each turn producing a colored ball, assuming the previous one was successful.

SVP
Joined: 30 Oct 2003
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Hell

I couldn't come out with 4/8 * 3/7 * 2/6 * 1/5
such a shame. This is the classical way of solving such problems.
Thanks to stoolfi for reminding me.
Director
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I wasn't trying to shame you, I want you to tell me how you came up with your answer, which, as I already wrote, is the same as mine.

I see people using combinatorics a lot when I don't use them, and I usually can't figure out why they are working...
SVP
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Well there is one ball of each color. So picking each with out replacement is 1/8 * 1/7 * 1/6 * 1/5
But you can arrange these 4 colored balls in 4! ways right
so you have 4! * 1/8 * 1/7 * 1/6 * 1/5

This is the same technique you use when you toss 5 coins and want to have exactly 3 Heads. You just find the probablity that 3 Hs will come and then multiply the P with no of combinations that give you 3Hs. This gives you total probability.

In the above example you treated 4 colored balls the same. This is perfect because the other balls are non colored. I liked your application of probability.
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