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audavidb
12-20-2002, 01:10 PM
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

happyjaypee
12-20-2002, 01:37 PM
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

irchans
12-20-2002, 01:44 PM

Chance of drawing 4 reds:

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

audavidb
12-20-2002, 01:47 PM
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

marbles
12-20-2002, 01:55 PM
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.

RocketManJames
12-22-2002, 04:56 AM
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