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Probability of marbles

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by Stockmoose16 » Wed Sep 17, 2008 2:51 pm
If you have 4 red marbles, 3 yellow, and 2 green, what is the probability of getting 2 Red, 2 Green, and 1 yellow, if the marbles aren't replaced:

I tried this two ways and got two different answers:

METHOD #1: Combinations

# of ways to get 2R, 2G, 1Y 2: 4C2*2C2*3C1 =18
Total # of ways to pick 5 balls: 9C5= 126

Probability: 18/126

METHOD #2: Fractions

4/9*3/8*2/7*1/6*3/5 = 72/15120
Multiply by 5 for the # of possible rearrangements (or is it 5!... not sure)

= 360/15120

Why are these numbers not even close to each other?
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Re: Probability of marbles

by pseudononymous » Wed Sep 17, 2008 6:05 pm
Stockmoose16 wrote:If you have 4 red marbles, 3 yellow, and 2 green, what is the probability of getting 2 Red, 2 Green, and 1 yellow, if the marbles aren't replaced:

I tried this two ways and got two different answers:

METHOD #1: Combinations

# of ways to get 2R, 2G, 1Y 2: 4C2*2C2*3C1 =18
Total # of ways to pick 5 balls: 9C5= 126

Probability: 18/126

METHOD #2: Fractions

4/9*3/8*2/7*1/6*3/5 = 72/15120
Multiply by 5 for the # of possible rearrangements (or is it 5!... not sure)

= 360/15120

Why are these numbers not even close to each other?
I think method #1 is correct. The 2nd method restricts the balls to a specific order (red has to be picked first).
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Re: Probability of marbles

by lunarpower » Thu Sep 18, 2008 1:10 am
pseudononymous wrote:
Stockmoose16 wrote:If you have 4 red marbles, 3 yellow, and 2 green, what is the probability of getting 2 Red, 2 Green, and 1 yellow, if the marbles aren't replaced:

I tried this two ways and got two different answers:

METHOD #1: Combinations

# of ways to get 2R, 2G, 1Y 2: 4C2*2C2*3C1 =18
Total # of ways to pick 5 balls: 9C5= 126

Probability: 18/126

METHOD #2: Fractions

4/9*3/8*2/7*1/6*3/5 = 72/15120
Multiply by 5 for the # of possible rearrangements (or is it 5!... not sure)

= 360/15120

Why are these numbers not even close to each other?
I think method #1 is correct. The 2nd method restricts the balls to a specific order (red has to be picked first).
actually, both methods are correct. you just have to multiply by the correct number in method #2; that number is neither 5 nor 5!. instead, that number is the number of different ways of organizing two r's, two g's, and one y, which is 5! / (2!2!1!) , or 30.

i posted more about this problem here.
Ron has been teaching various standardized tests for 20 years.

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by 4meonly » Thu Sep 18, 2008 6:17 am
somebode, what is OA?

I have 1/7

(4/9*3*8*2/7*1/6*3/5)*5!/2!*2! = 1/7
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by lunarpower » Fri Sep 19, 2008 2:28 am
4meonly wrote:somebode, what is OA?

I have 1/7

(4/9*3*8*2/7*1/6*3/5)*5!/2!*2! = 1/7
the answer to the problem is definitely 1/7, as mentioned in the thread linked above.
Ron has been teaching various standardized tests for 20 years.

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