A complete guide to Permutation, Combination and Probability. All you need for GMAT

[TwinnSplitter]

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.

[prometheus]

good catch. thanks

[lego2401]

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?

[TwinnSplitter]

personally I just take the adhoc approach, 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.

[lego2401]

Originally Posted by TwinnSplitter

personally I just take the adhoc approach, 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 .........

[TwinnSplitter]

Originally Posted by lego2401

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 .........

haha yup that's what I meant...I think a lot of the people on this site are just so used to knowing a great deal of math, so when they see ridiculously hard combination problems and can't do them, they feel as though they have to learn absolutely everything there is to know about them.

[dadaruh]

Man, how about #3 of probability questions?

I keep finding 27/28 however, you found 15/28 how come?

appriciate your reply
thx

ugur

[TwinnSplitter]

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?



A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

I believe you interpretted the question to mean that what is the probability that either 0 or 1, but not 2, of the marbles chosen is blue, in which case 27/28 would be correct.

However, I think the question is asking what's the probability that NONE of the marbles chosen are blue. Then we get
(6/8) x (5/7) = 15/28

[dadaruh]

Hey, since I recognized the answer you have given to Lego's similar question, i got the way you answered to #3. However, regarding the question, I thought that none-blue can be found if we dare to find the all-of-them-blues and subtract from 1. It seems wrong approach but, could you give a reason?


i appriciate.

ugur

[TwinnSplitter]

Basically, you want to use the way that will allow you to use the least number of probability calculations. So, for example, if it asks "what is the probability that you get at least one blue", then the options are that you get 1 blue or you get 2 blues. However, instead of calculating each of these individual probabilities, it is easier to calculate
1 minus the probability of getting 0 blues.

But in the case where it says what is the probability of getting 0 blues, then it's easiest to just use the probability of getting two "not blues".

And if it asked you what is the probability of 2 blues, then it would be easiest to just find it as (2/8) x (1/7) = 1/28.

In other words, just think which way will take you the smallest number of probability caclulations.

Hope that helps.