What is wrong with my approach ?

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Sat Aug 27, 2011 8:50 pm

by [email protected] » Sun Mar 18, 2012 9:25 am

I am having trouble in getting the right answer using my approach for the below question. Please help.

A bag holds 4 red marbles, 5 blue marbles, 2 green marbles. If 5 marbles are selected one after the other without replacement, what is the probability of drawing 2 red marbles, 2 blue marbles and 1 green marble ?

I solved it as below but this is not matching the answer.

4/11 * 3/10 * 5/9 * 4/8 * 2/7 = 2/231

Can anyone please check ?

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Source: Beat The GMAT — Problem Solving | Sun Mar 18, 2012 9:25 am

jzw: Senior | Next Rank: 100 Posts; Posts: 79; Joined: Mon Feb 13, 2012 3:02 pmFollowed by: 3 members; Thanked: 2 times

by jzw » Sun Mar 18, 2012 10:01 am

Can you please make sure that the question you wrote below is using the same exact text as it appeared and also include the answer choices. Thanks.

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Sat Aug 27, 2011 8:50 pm

by [email protected] » Sun Mar 18, 2012 10:22 am

The text is the same. This is from KAPLAN GMAT PREMIER 2011 - page 260.

This is an example question and does not have any options. They have provided an approach to solve this which is different than mine. Since the answers are not matching, i wanted to know the flaw in my approach.

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seal4913: Master | Next Rank: 500 Posts; Posts: 106; Joined: Wed Mar 23, 2011 4:56 pm; Thanked: 8 times

by seal4913 » Sun Mar 18, 2012 11:17 am

Your way you are saying order matters. here it doesn't thats why it's a combination question. If they asked what is the extact way to a certain outcome your apporach would be ok. So for any prob question it's the # of desirble outcomes/over the total # of outcomes.

So the total number of outcomes is 11C5. So that 11!/[5!(11-5)!] which is 11 x 3 x 2 x 7

Then you have to find out the combos of finding each color.

So blue is 5C2, red is 4C2 and green is 2C1.

So that's 5!/(2!3!) = 10
Then 4!/(2!2!) = 6
Then 2!/(1!1!) = 2

So the number of desirble outcomes is 10 x 6 x 2 divided by total 11 x 3 x 2 x 7 = 10 x 2/11 x 7

which is 20/77

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jzw: Senior | Next Rank: 100 Posts; Posts: 79; Joined: Mon Feb 13, 2012 3:02 pmFollowed by: 3 members; Thanked: 2 times

by jzw » Sun Mar 18, 2012 11:34 am

Ok - here is what I did and why I did it. Please confirm whether the answer is correct. This question is different than other probability questions in that it is also a combination problem.

The first thing we need to do in any probability question is to find out what are the desired outcomes and put that over the total possible outcomes. Desired/Total. But what's the total? How many possible marbles could we pull out? Well, there are 11 marbles, but we'll only be picking 5 of them. That's going to be 11C5, which ends up being 77. Ok, so that's the denominator. But now, we need to calculate the top portion, which is to figure out what are the chances of choosing those colored marbles.

The order you are pulling out the marbles in does not seem to matter, which means we will need to divide something. (order matters = don't divide, order does no matter = divide).

So this means, we need think - how many marbles do we want for each selection and what are our total slots available for each colored marble. Well, it says we need 2 red marbles, and there are 4 red marbles. That's gunna be 4C2. We need 2 blue marbles and there are 5, so that's 5C2, and finally we need green marbles and there are 2, and that's 2C1.

So now we've gotta take 4C2 x 5C2 x 2C1 and multiply that sucker out, which becomes 22.

So now we have our numerator 22 and stick it over the denominator, which we established is 77.

So - my answer for this one is 22/77.

Anywhere near the correct answer?

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seal4913: Master | Next Rank: 500 Posts; Posts: 106; Joined: Wed Mar 23, 2011 4:56 pm; Thanked: 8 times

by seal4913 » Sun Mar 18, 2012 11:36 am

jzw wrote:Ok - here is what I did and why I did it. Please confirm whether the answer is correct. This question is different than other probability questions in that it is also a combination problem.

The first thing we need to do in any probability question is to find out what are the desired outcomes and put that over the total possible outcomes. Desired/Total. But what's the total? How many possible marbles could we pull out? Well, there are 11 marbles, but we'll only be picking 5 of them. That's going to be 11C5, which ends up being 77. Ok, so that's the denominator. But now, we need to calculate the top portion, which is to figure out what are the chances of choosing those colored marbles.

The order you are pulling out the marbles in does not seem to matter, which means we will need to divide something. (order matters = don't divide, order does no matter = divide).

So this means, we need think - how many marbles do we want for each selection and what are our total slots available for each colored marble. Well, it says we need 2 red marbles, and there are 4 red marbles. That's gunna be 4C2. We need 2 blue marbles and there are 5, so that's 5C2, and finally we need green marbles and there are 2, and that's 2C1.

So now we've gotta take 4C2 x 5C2 x 2C1 and multiply that sucker out, which becomes 22.

So now we have our numerator 22 and stick it over the denominator, which we established is 77.

So - my answer for this one is 22/77.

Anywhere near the correct answer?

Close 22 should be 20 see above.

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jzw: Senior | Next Rank: 100 Posts; Posts: 79; Joined: Mon Feb 13, 2012 3:02 pmFollowed by: 3 members; Thanked: 2 times

by jzw » Sun Mar 18, 2012 11:46 am

err - whoops. or - oh yeah, I did that on purpose to see if you were paying attention

that's exactly the kind of simple, bumbling, idiotic, stupid, error I'll make on test day that could totally trip me up. it sucks when you know how to do it and then get the wrong answer because of some ridiculous mistake. sometimes i get lucky and if i get an answer that's nowhere in the answer choices i realize that i made what i call a "simple error" and i can usually trace it back quickly and correct. but sometimes they predict the error and stick that one in as one of the answer choices.

lesson - don't rush. better to work slower and get the correct answer the first time than having to go back and correct.

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Sat Aug 27, 2011 8:50 pm

by [email protected] » Sun Mar 18, 2012 1:43 pm

I understood the problem with my approach now.
Thanks a lot guys!!

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

with

by pemdas » Thu Mar 22, 2012 2:29 pm

this question isn't precise and must be reworded for the answer 20/77 with the solution proceeding from combined counting 4C2*5C2*2C1/11C5. If the color is the only difference amongst the marbles listed and all other attributes are alike, then yes it's 20/77; under the given constraints we cannot solve this question as there is an insufficiency of data introduced - not all attributes are clear.

seal4913 wrote:Your way you are saying order matters. here it doesn't thats why it's a combination question. If they asked what is the extact way to a certain outcome your apporach would be ok. So for any prob question it's the # of desirble outcomes/over the total # of outcomes.

So the total number of outcomes is 11C5. So that 11!/[5!(11-5)!] which is 11 x 3 x 2 x 7

Then you have to find out the combos of finding each color.

So blue is 5C2, red is 4C2 and green is 2C1.

So that's 5!/(2!3!) = 10
Then 4!/(2!2!) = 6
Then 2!/(1!1!) = 2

So the number of desirble outcomes is 10 x 6 x 2 divided by total 11 x 3 x 2 x 7 = 10 x 2/11 x 7

which is 20/77

Success doesn't come overnight!

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seal4913: Master | Next Rank: 500 Posts; Posts: 106; Joined: Wed Mar 23, 2011 4:56 pm; Thanked: 8 times

by seal4913 » Thu Mar 22, 2012 2:39 pm

If you read what he writes it comes from the Kaplan book so I looked it up. I got what the question was and clearly explained what he was doing wrong and what the correct answer is. There is sufficent data.

pemdas wrote:this question isn't precise and must be reworded for the answer 20/77 with the solution proceeding from combined counting 4C2*5C2*2C1/11C5. If the color is the only difference amongst the marbles listed and all other attributes are alike, then yes it's 20/77; under the given constraints we cannot solve this question as there is an insufficiency of data introduced - not all attributes are clear.
seal4913 wrote:Your way you are saying order matters. here it doesn't thats why it's a combination question. If they asked what is the extact way to a certain outcome your apporach would be ok. So for any prob question it's the # of desirble outcomes/over the total # of outcomes.

So the total number of outcomes is 11C5. So that 11!/[5!(11-5)!] which is 11 x 3 x 2 x 7

Then you have to find out the combos of finding each color.

So blue is 5C2, red is 4C2 and green is 2C1.

So that's 5!/(2!3!) = 10
Then 4!/(2!2!) = 6
Then 2!/(1!1!) = 2

So the number of desirble outcomes is 10 x 6 x 2 divided by total 11 x 3 x 2 x 7 = 10 x 2/11 x 7

which is 20/77

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Thu Mar 22, 2012 2:53 pm

never mind - i didn't mean to complicate the issues. perhaps we are apart on this subject, as i looked this entry as one missing info on quantitative and qualitative attributes not introduced here for each of 5-marble, 4-marble and 2-marble sets of the same color(s). If the question would mention that there is not any other difference amongst marbles in the given sets of the same color, that is counting-wise P(selecting 1 marble of red color) is always 4C1* .../11C5 or P(selecting 2 marbles of red color) will be 4C2*.../11C5 - possible if a color is the only qualitative attribute different for the given set - then yes ...
anyways, it's beyond that of Kaplanish method of teaching GMAT and of course won't be studied/exercised outside of statistics class. The thing is that official GMAT sources are always precise in listing all constraints to avoid inaccuracy, even with questions having statistical implications.

Success doesn't come overnight!

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