lego2401 Posted April 19, 2005 Share Posted April 19, 2005 Hey guys, I came across this site a couple of nights ago and after reviewing it, found that it would be helpful to anyone taking the GMAT. The site pretty much wraps up all the PERM/COMB/PROB. concepts you'll need for the GMAT or that have been discussed in this forum. Hope it helps..... http://www.mansw.nsw.edu.au/members/reflections/vol23no4yen.htm PS: Please keep in mind that the GMAT is not a Discrete Maths exam, ie, you will see no more than 3-4 of these question on the exam, and thats if ur doing really well, so don't spend too much time on this topic. 4 Link to comment Share on other sites More sharing options...
greycellz Posted April 19, 2005 Share Posted April 19, 2005 I too have an interesting document that some good soul prepared. See if any of you find that helpful:Probability.doc 4 Link to comment Share on other sites More sharing options...
prep2max Posted April 19, 2005 Share Posted April 19, 2005 Thanks Lego and Grey! great job:tup: These pretty much sum up everything one needs to know regarding the respective(dreaded) topics. grey where are the answers to the probability questions or can we post the answers to all and then you can give out the OA's..like a mini-test:) Link to comment Share on other sites More sharing options...
NICOLASSW1 Posted April 19, 2005 Share Posted April 19, 2005 That was a great article!!! It helped me alot. Link to comment Share on other sites More sharing options...
Mattt Posted April 20, 2005 Share Posted April 20, 2005 I would also like the answers to the probability questions Link to comment Share on other sites More sharing options...
greycellz Posted April 21, 2005 Share Posted April 21, 2005 Thanks Lego and Grey! great job:tup: These pretty much sum up everything one needs to know regarding the respective(dreaded) topics. grey where are the answers to the probability questions or can we post the answers to all and then you can give out the OA's..like a mini-test:) No OAs, these are as is. We can discuss if you like.. Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 21, 2005 Share Posted April 21, 2005 Thanks for these great resources. My answers to that document are as follows, anybody else who did them let me know what you got: B C A C D (the way I did this took kind of long, if somebody else could let me know how they did it I'd appreciate it) B B E E D B D E E B E A D D Link to comment Share on other sites More sharing options...
Anandha Posted April 21, 2005 Share Posted April 21, 2005 The articles are very good and helpful. Thank you to both of you. Link to comment Share on other sites More sharing options...
mally Posted April 21, 2005 Share Posted April 21, 2005 its sad but the link is gone to http://www.mansw.nsw.edu.au/members...vol23no4yen.htm its no longer there.:( Link to comment Share on other sites More sharing options...
lego2401 Posted April 21, 2005 Author Share Posted April 21, 2005 its sad but the link is gone to http://www.mansw.nsw.edu.au/members...vol23no4yen.htm its no longer there.:( Yeah it is, I just tried it. Link to comment Share on other sites More sharing options...
mally Posted April 21, 2005 Share Posted April 21, 2005 I'm sorry-that means somethings wrong on my comp! Link to comment Share on other sites More sharing options...
mally Posted April 21, 2005 Share Posted April 21, 2005 Got it on this one. Link to comment Share on other sites More sharing options...
mally Posted April 21, 2005 Share Posted April 21, 2005 Hi, Thanks for your answers , I'm solving the exercise and comparing my answers with yours.How did you do No.4-can u show me? your answer is D isn't it? I don't know why B is marked already. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue? A. 21/50 B. 3/13 C. 47/50 D. 14/15 E. 1/5 Thanks! Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 21, 2005 Share Posted April 21, 2005 probability that they're both red = (3/10) x (2/9) = 1/15 Probability that at least one is blue = 1 - probability that they're both red = 1 - 1/15 = 14/15 So the answer is D Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 21, 2005 Share Posted April 21, 2005 were the rest of your answers the same as mine? Link to comment Share on other sites More sharing options...
mally Posted April 21, 2005 Share Posted April 21, 2005 My answer for the 9th problem is D not E.How did u arrive at 1/32? Problem given below: If four fair coins are tossed, what is the probability of all four coming up heads? A. 1/4 B. 1/6 C. 1/8 D. 1/16 E. 1/32 Link to comment Share on other sites More sharing options...
mally Posted April 22, 2005 Share Posted April 22, 2005 nope.look at the next one i've posted about 4 coins. But I'll post all my answers -still going through it.whew! Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 22, 2005 Share Posted April 22, 2005 I also put D (not E) for that question. You were probably looking at the two E's that I put for the two questions before that. So I got 1/16 too:) Link to comment Share on other sites More sharing options...
Nity Posted April 22, 2005 Share Posted April 22, 2005 Thanks for the Probability document and the link. They've been really helpful Link to comment Share on other sites More sharing options...
prometheus Posted April 24, 2005 Share Posted April 24, 2005 Do we have the OAs ? My answers: B,C ,A,C, D,B,B,E,E,D,B,D,E,D,B,E,A,D,D,D Twinsplitter: My answers were the same as yours except for the 6th question from the end. A fair, six-sided die is rolled. What is the probability of obtaining a 3 or an odd number? My answer to this was 1/6 + 1/2 = 2/3 and option (D). Is this right ? Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 24, 2005 Share Posted April 24, 2005 Hi prometheus, the trick in this question is that 3 IS an odd number, so they're trying to get you to double count it. The probability of getting a 3 is already included in calculating the probability of getting an odd number, and thus the answer is 1/2. Hope that makes sense. Link to comment Share on other sites More sharing options...
prometheus Posted April 24, 2005 Share Posted April 24, 2005 good catch. thanks Link to comment Share on other sites More sharing options...
lego2401 Posted April 24, 2005 Author Share Posted April 24, 2005 Point for discussion: From what I've seen on the forum in the last while, the sticky parts in probability are, 1-when to use combinations vs not to, or when they can be used interchangebly, and 2- When asked to find the probability of "atleast" a certain event, when to double count and when not to: Here are examples of the second point: A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue? In this instance, we take 3 situations, namely where the first is blue, second not blue, or first is not blue, second is not blue, or both are blue. Now in the second example, A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue? In this instance, we approach the problem by counting the sitation where none are blue, that is, both red. The point I see mistakes in is, when the two are of different colors, we count twice, ie, we consider RB then BR, whereas when the two are of the same color, we count only once. Let's discuss these points and see if anyone has some kind of fixed rather than adhoc approach. Finally, problem to work with (this is a variation of one of the problems in Grey's document): Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that NONE of them are blue? Link to comment Share on other sites More sharing options...
TwinnSplitter Posted April 24, 2005 Share Posted April 24, 2005 personally I just take the adhoc approach:blush:, as I haven't really come to have a fixed, step by step strategy. I just look at the problem, try and understand what it's asking, and execute accordingly. As for the question you posted, I would divide them up into two groups: blue and not blue (not blue being red and green). The first time you pick, the probability of getting "not blue" is 4/7, and the second time you pick, the probability of getting "not blue" (assuming you got "not blue" the first time) is 3/6. The third time has probability 2/5. Thus, the probability of getting no blues is: (4/7) x (3/6) x (2/5) = 4/35 I do agree with you that these questions are on the level of what you'll see on the GMAT, and I think that a lot of people are spending way too much time on extremely difficult probability and not enough time on the more fundamental aspects of GMAT math. Link to comment Share on other sites More sharing options...
lego2401 Posted April 25, 2005 Author Share Posted April 25, 2005 personally I just take the adhoc approach:blush:, as I haven't really come to have a fixed, step by step strategy. I just look at the problem, try and understand what it's asking, and execute accordingly. As for the question you posted, I would divide them up into two groups: blue and not blue (not blue being red and green). The first time you pick, the probability of getting "not blue" is 4/7, and the second time you pick, the probability of getting "not blue" (assuming you got "not blue" the first time) is 3/6. The third time has probability 2/5. Thus, the probability of getting no blues is: (4/7) x (3/6) x (2/5) = 4/35 I do agree with you that these questions are on the level of what you'll see on the GMAT, and I think that a lot of people are spending way too much time on extremely difficult probability and not enough time on the more fundamental aspects of GMAT math. I hope you meant there that you agree with me that these questions are NOT on the level of what'll be seen on the GMAT :) I have being saying that for a while, prob will be atmost a couple of question, and theyll be easy ones. I think people have to focus on other points and spend less time on this topic ......... Link to comment Share on other sites More sharing options...
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