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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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 |
Re: Math Question
I got a different answer.
Chance of drawing 4 reds: 14/16*13/15*12/14* 11/13 = 11/20 = 55% |
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 |
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. |
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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