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  #1  
Old 12-20-2002, 01:10 PM
audavidb audavidb is offline
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Default Math Question

If you have 16 marbles in a bag and 14 are red and 2 are blue. You pull 4 out, what is the likely hood that none pulled will contain a blue marble. Thanks, David
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  #2  
Old 12-20-2002, 01:37 PM
happyjaypee happyjaypee is offline
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Default Re: Math Question

Number of total possible draws:

16 X 15 X 14 X 13 = 43,680


Number of draws containing at least a blue marble:

2 X 15 X 14 X 13 = 5,460


Probability of drawing at least a blue marble:

5,460 X 100 / 43,680 = 12.5%


Probability of drawing four red:

100% - 12.5% = 87.5% or odds of 7:1
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  #3  
Old 12-20-2002, 01:44 PM
irchans irchans is offline
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Default Re: Math Question

I got a different answer.

Chance of drawing 4 reds:

14/16*13/15*12/14* 11/13 = 11/20 = 55%

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Old 12-20-2002, 01:47 PM
audavidb audavidb is offline
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Default Re: Math Question

Thanks for the response, but I think its incorrect. I think I have it figured out....

16X15X14X13/4X3X2X1=1820 Total outcomes possible
14X13X12X11/4X3X2X1=1001 Desired outcomes (no blue)

55% chance of no blue, 45% of a blue. If anyone thinks this is wrong, please let me know.

Thanks
DAvid
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  #5  
Old 12-20-2002, 01:55 PM
marbles marbles is offline
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Default Re: Math Question

Your result is correct. I take what I believe to be a simpler approach, taking the balls out one at a time, trials 1-4.
Trial 1: 14/16
Trial 2: 13/15
Trial 3: 12/14
Trial 4: 11/13
Multiply the 4 numbers together, and you reach the same 0.55 probability.
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  #6  
Old 12-22-2002, 04:56 AM
RocketManJames RocketManJames is offline
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Default Re: Math Question

Yet another approach (which could be considered even simpler, depending on familarity with Choose).

There are 16 C 4 possible draws. There are 14 C 4 possible draws excluding the blues.

(14 C 4) / (16 C 4) reduces to 11*12/15*16 = 55%.

Just another way to look at it.

-RMJ
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