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If you have 250 marbles, 15 are black, and you pull 10 at

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Manager
Joined: 13 Feb 2007
Posts: 63
If you have 250 marbles, 15 are black, and you pull 10 at [#permalink]

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21 May 2007, 16:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If you have 250 marbles, 15 are black, and you pull 10 at random, each day for 5 days (the marbles are then returned before the next selection), what is the probability that one of the black marbles will not be grabbed?
Manager
Joined: 28 Aug 2006
Posts: 160

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21 May 2007, 19:55
Probability of getting a blcak marble is 15/250=3/50

Probablity of not getting black is 47/50

Probability of 10 marbles selected with no black

10C0 (3/50)^0 (47/50)^10

SInce the same is repeated 5 days it is

Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

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21 May 2007, 23:21
5 x (235C10/250C10)
Senior Manager
Joined: 04 Mar 2007
Posts: 434

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22 May 2007, 11:26
vijay2001 wrote:
Probability of getting a blcak marble is 15/250=3/50

Probablity of not getting black is 47/50

Probability of 10 marbles selected with no black

10C0 (3/50)^0 (47/50)^10

SInce the same is repeated 5 days it is

As far as I understood the problem asks NOT about the prob. of not selecting any black marbles, but about the probability of selecting only 14 out of 15 black marbles....

I think it must be:
14C50*(3/50)^14*(1-3/50)^36
Senior Manager
Joined: 04 Mar 2007
Posts: 434

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24 May 2007, 22:29
Hey guys,
Can anyone explain it to me please?
What is the right solution?

Thanks!
VP
Joined: 08 Jun 2005
Posts: 1145

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25 May 2007, 14:02
Caas wrote:
Hey guys,
Can anyone explain it to me please?
What is the right solution?

Thanks!

In order to solve and better understand the above problem, I will start with a similar, but yet easier, problem (if you think it's too easy skip it).

Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red?

A.2/5
B.6/10
C.7/10
D.15/20
E.19/20

to solve this problem we need only to apply logic (no combinatorics or probability formulas needed !)

5 = total marbles
3 = blue marbles
2 = red marbles

Andy picks 2 marbles, and we want to know what is the probability that he will pick at least one red marble (note it can be two or one).

2/5 (first pick, probability of choosing red is two out of five)
1/4 (second pick, probability of choosing red is one out of four since one was removed in first pick).

Since the events are dependent we need multiply them.

2/5*1/4 = 2/20 = 1/10 (probability of choosing red in first pick and choosing red in second pick)

but as you can see thats not even one of the given choices !!

we need to add outcomes where Andy chooses only one red !

2/5*3/4 = 6/20 = 3/10 (choosing red in the first pick and choosing blue in the second pick)

&

3/5*2/4 = 6/20 = 3/10 (choosing blue in the first pick and choosing red in the second pick)

since those three events are independent we will add them one to another.

(3/10)+(3/10)+(1/10) = 7/10

Moveing on to the posted question:

If you have 250 marbles, 15 are black, and you pull 10 at random, each day for 5 days (the marbles are then returned before the next selection), what is the probability that one of the black marbles will not be grabbed?

My opinion is that this is way too complex question to be in the GMAT. but nevertheless applying the same logic as in the first problem will help us solve this problem:

250 = total marbles
15 = black marbles
235 = not-black marbles (we really don't care what color).

the probablity of choosing a not-black marble in a single try is P1 = 235/250 = 94%

the probablity of choosing a not-black marble in ten tries is

P10 = (235/250)*(235/250)*(235/250)*(235/250)*(235/250)*(235/250)*(235/250)*(235/250)*(235/250)*(235/250) = (235/250)^10 = ~ 54%

choosing not-black in first,second,third... and so on pick, for a total of ten picks)

we will use (235/250) every pick (and not 234/250, 233/250) because the marbles are being returned after each selection (unlike the first problem).

Since we have the same process repeted over five days, and the events are dependent (statistically it dosen't matter if we pick 50 times in one day or pick 10 times a day for five days !).

we have to multiply (235/250) 50 times !!!

The outcome will then be (235/250)^50 = ~ 4.5%

we have ~ 4.5% of chooseing not-black every single time (meaning not choosing black marble) marble !

Manager
Joined: 28 Aug 2006
Posts: 160

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27 May 2007, 21:28
This question has a lot of BS....

Total number of marbles =250
Number of non black marbles =235

Probability of choosing 10 non black marbles is 235C10/250C10

This is repeated 5 times. Hence the probability for not picking a black marble for 5 consequtive days is (235C10/250C10) ^ 5

PS: Hey killer, yours would be true if 10 marbles were picked one at a time.
Director
Joined: 03 Sep 2006
Posts: 871

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27 May 2007, 21:48
5 x (235C10/250C10)

In 60 seconds, I could come up with same solution as yours.

Is this the OA??
27 May 2007, 21:48
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