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Apr 2808:00 AM PDT
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16 Mar 2014, 18:12
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A bag holds 4 red marbles, 5 blue marbles, and 2 green marbles. If 5 marbles are selected one after another without replacement, what is the probability of drawing 2 red marbles, 2 blue marbles, and 1 green marble?
My question is, why can't I do: (4/11) x (3/10) x (5/9) x (4/8) x (2/7) ?
I feel like I've seen several similar questions that do the method above. Why wouldn't it work here? Is what I did above showing the probability of selecting 2 red, 2 blue, and 1 green in that exact order?
Kaplan has 20/77 as the answer… Any help would be much appreciated. Thanks.
My question is, why can't I do: (4/11) x (3/10) x (5/9) x (4/8) x (2/7) ?
I feel like I've seen several similar questions that do the method above. Why wouldn't it work here? Is what I did above showing the probability of selecting 2 red, 2 blue, and 1 green in that exact order?
Kaplan has 20/77 as the answer… Any help would be much appreciated. Thanks.
17 Mar 2014, 10:57
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Rtr6262
Dear Rtr6262,
I'm happy to respond.
First of all, this question would probably be better posted in the Quant section, because it's an individual math question. The "ask GMAT Experts" section is typically for strategy and more general questions, not for individual test question.
Probability is hard --- it takes a lot of practice to develop the worldview, to understand exactly what is being calculated in each situation. Many students naively focus simply on "what to do" --- while that is somewhat problematic in all branches of math, it's a HUGE problem in probability. It's very important to focus on the worldview, to focus primarily on "how to see", that is, how to organize and interpret information given in a problem.
You are perfectly correct --- what you calculated was the probability of picking that combination in a specific order. Of course, what the question wants is the probability of winding up with those five balls, but we don't care at all about the order in which they are picked.
I would think about this question in terms of counting methods. That's one of many approaches. See:
https://magoosh.com/gmat/2013/gmat-proba ... echniques/
Suppose we number all eleven balls. Then, the total number of combinations of five we could pick would be
11C5 = (11!)/[(6!)*(5!)] = (11*10*9*8*7)/(5*4*3*2*1) = (11*9*8*7)/(4*3)
= (11*9*2*7)/(3) = 11*3*2*7
I'm just going to leave that in unmultiplied form. That's our denominator.
For our numerator, we want groups with two reds, two blues, and greens.
For the two reds, we are picking two of 4, so that's 4C2 = 6 possibilities
For the two blues, we are picking two of 5, so that's 5C2 = 10 possibilities
For the one green, we have two possibilities.
By the Fundamental Counting Principle, we multiply those for our numerator --- 6*10*2 --- again, I will leave that in unmultiplied form.
Now, our probability fraction is:
(6*10*2)/(11*3*2*7) = (10*2)/(11*7) = 20/77
That's the OA.
Does this approach make sense? This is not the only way to solve this, but solutions by counting techniques is a powerful trick to have up your sleeve.
Let me know if you have any further questions.
Mike
17 Mar 2014, 21:05
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Rtr6262
You can but what you have calculated is the probability of picking RRGGB. What about cases such as RGRGB or BGRRG etc
How many such cases are possible? You can arrange 5 objects in 5! ways. Since 2 pairs are of the same color, you get 5!/2!*2! = 30 arrangements.
(4/11) x (3/10) x (5/9) x (4/8) x (2/7) * 30 = 20/77
