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Old 2009 March 18th, 08:52 AM   #1 (permalink)
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A very challenging probability problem from 800 score.com

http://img25.imageshack.us/img25/842...00914724pm.png
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Old 2009 March 18th, 02:38 PM   #2 (permalink)
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Dear Kesho,
It is really a great problem. is the answer 7?

Would you give me the referrence form where you have taken it.

Thanking you-
smbabu
sabyasachymistry@yahoo.com
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Old 2009 March 18th, 04:16 PM   #3 (permalink)
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I also concur with 7 because the worst case sequence is R G B R G B
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Old 2009 March 19th, 05:09 AM   #4 (permalink)
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Yes the answer is 7.

Please explain in details.
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Old 2009 March 19th, 09:35 AM   #5 (permalink)
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Dear Kesho i believe that the way to get the answer is as follows:
No. of marbles = 18
No. of Red marbles = 8
No. of Green marbles = 4
No. of Black marbles = 6
Now,
The probability of picking up 3 red's = 1/R = 3/8
The probability of picking up 3 greens = 1/G = 3/4
The probability of picking up 3 blacks = 1/B = 3/6 or 1/2

Hence the probability of picking up 3 of the same colour is =

1/P = 1/R*1/G*1/B
= 3/8*3/4*1/2
= 9/64

Finally,
1/P = 9/64
P = 64/9
P = 7.111111 ~ 7
Hence, P = 7

Hope this answers the Q?

Please do correct me if i am wrong.
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Old 2009 March 19th, 02:49 PM   #6 (permalink)
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Quote:
Originally Posted by SuperFlanker View Post
I also concur with 7 because the worst case sequence is R G B R G B
Understood.

@ MTAB1380, superflanker, smbabu

thanks
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Old 2009 March 19th, 07:16 PM   #7 (permalink)
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Dear kesho,
Thank you very much. is 'MTAB 1380' available at internet?
-
smbabu
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Old 2009 August 2nd, 10:48 PM   #8 (permalink)
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i disagree, the question isn't asking about probability, it's asking about the total number you have to pick before getting at least one of each color

Answer should be D 13

Assume the worst case scenario if he picks only red and black that's 12 marbles

he would only need to pick one more to get one of each color
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Old 2009 August 2nd, 11:47 PM   #9 (permalink)
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Quote:
Originally Posted by dknight237 View Post
Assume the worst case scenario if he picks only red and black that's 12 marbles
only red and black will make 14 marbles!
so i think the answer is E
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Old 2009 August 3rd, 04:11 AM   #10 (permalink)
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dknight237, you probably misunderstood the question. It asks what's the minimum number to draw to have at least 3 of the same color -- not one of each color.

Like SuperFlanker said, the worst case scenario is to have two of each color (six picks). The student will have at least three of one color on his 7th pick.
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