The confusion over order in probability...

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Stockmoose16: Master | Next Rank: 500 Posts; Posts: 347; Joined: Mon Aug 04, 2008 1:42 pm; Thanked: 1 times

The confusion over order in probability...

by Stockmoose16 » Wed Sep 17, 2008 7:15 pm

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There is a lot of confusion on this board over probability and whether order matters when calculating the answer. I'm hoping a GMAT expert can weigh in on the difference between the following two problems. In Case #1, it appears that ORDER DOES MATTER. And in case #2, ORDER DOES NOT seem to matter. Can someone explain why there's a difference?

Question #1: A bag has 6 red marbles and 4 marbles. What are the chances of pulling out a red and blue marble.

Answer: (Chance of picking red * chance of picking blue) + Chance of picking blue*chance of picking red)

6/10*4/9 + 4/10*6/9 = 48/90

***Here, it appears that order matters, because you have to find the probability of getting red-blue, and add it to the probability of getting blue-red.

Question #2:
A bag has 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:

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

Answer: 18/126

Now here, it appears order DOES NOT matter, since we're using combinations (instead of permutations).

How come in Q#1, blue-red, red-blue are distinct, which means you have to add the probabilities of each together, but in Q#2, it uses combinations, which means order doesn't matter.

Hopefully an expert can weigh in, because I'm sure plenty of people are confused by this.

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Source: Beat The GMAT — Problem Solving | Wed Sep 17, 2008 7:15 pm

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 am; Thanked: 2256 times; Followed by:1535 members; GMAT Score:800

Re: The confusion over order in probability...

by lunarpower » Thu Sep 18, 2008 12:39 am

Stockmoose16 wrote:There is a lot of confusion on this board over probability and whether order matters when calculating the answer. I'm hoping a GMAT expert can weigh in on the difference between the following two problems. In Case #1, it appears that ORDER DOES MATTER. And in case #2, ORDER DOES NOT seem to matter. Can someone explain why there's a difference?

Question #1: A bag has 6 red marbles and 4 marbles. What are the chances of pulling out a red and blue marble.

Answer: (Chance of picking red * chance of picking blue) + Chance of picking blue*chance of picking red)

6/10*4/9 + 4/10*6/9 = 48/90

***Here, it appears that order matters, because you have to find the probability of getting red-blue, and add it to the probability of getting blue-red.

Question #2:
A bag has 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:

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

Answer: 18/126

Now here, it appears order DOES NOT matter, since we're using combinations (instead of permutations).

How come in Q#1, blue-red, red-blue are distinct, which means you have to add the probabilities of each together, but in Q#2, it uses combinations, which means order doesn't matter.

Hopefully an expert can weigh in, because I'm sure plenty of people are confused by this.

the problem is the false dichotomy: there's no one correct way to do either of these problems. either one of them can be solved in either way.
effectively, order doesn't matter - but you can ALWAYS formulate problems in which order doesn't matter in some way in which it does. there's no damage done by considering order when it's unnecessary - all that happens is that both the total # of outcomes and the # of successes are multiplied by some constant (a constant that equals the number of times each non-ordered set is repeated when order starts to matter).

the first problem, for instance, can also be formulated in a way such that "order" doesn't matter:
# of ways of picking 1 red and 1 blue marble = (6c1)(4c1) = 6 x 4 = 24;
# of ways of picking 2 marbles in general = 10c2 = 45.
therefore, probability = 24/45. note that this is the same number arrived at by the consecutive multiplication technique above.
and note that 24 is half of 48, and 45 is half of 90 - because, when you consider order, each un-ordered set now appears twice.
both are correct.

also, as a takeaway, note that order is automatically assumed to matter when you use consecutive multiplication, so you must find all possible orders in which the desired event can occur. this is why the original solution for problem #1 features a product for "red then blue" and another product for "blue then red".

--

you could also formulate the second problem in a way such that order does matter, but you'd have an obscene number of products to add up. in particular, you'd have a separate product for each of the following:
rrggy
rrgyg
rrygg
rgrgy
rgryg
rgygr
...
...
yggrr
(5!/2!2!1! = 30 products in all)
each product has a numerator of (4 x 3) x (3) x (2 x 1) (for red, yellow, and green, respectively), and a denominator of (9 x 8 x 7 x 6 x 5) because there's no replacement.
that's 1/210 for each fraction, times thirty = 1/7.
note that 18/126 is also 1/7.
booya!

--

general takeaway:
whatever opener you use for a probability problem is likely to work, although some openers are way more work than others.

Ron has been teaching various standardized tests for 20 years.

--

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On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 am; Location: Montreal; Thanked: 1090 times; Followed by:355 members; GMAT Score:780

by Ian Stewart » Thu Sep 18, 2008 4:38 am

I also discussed whether order matters in these types of probability questions at the end of this thread:

www.beatthegmat.com/combination-and-per ... t8924.html

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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tnaim: Senior | Next Rank: 100 Posts; Posts: 67; Joined: Tue Mar 16, 2010 2:06 pm; Thanked: 3 times

by tnaim » Thu Jul 29, 2010 7:09 pm

Ron, Ian
The above two posts are much appreciated. I am glad to have found this thread and I would really appreciate your help in understanding a few more things as I have spent quite a bit of time trying to figure them out but with no luck.
1) If I understand you correctly, did you say in the posts above that we can pretty much solve any probability problem either way: by considering that order matters or by considering that order does NOT matter as long as we are consistent in applying our pick to both the number of favorable outcomes and the number of possible outcomes? did I get that right? if I have, I think that the implications are very important and one of them is: I no longer have to think whether order matters in any probability problem. I can simply learn how to solve any probability problem by considering that order does NOT matter and just make sure I apply that to both the number of favourable and possible outcomes
2) A case in point is this question: A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?
A) Â¼
b)3/8
c)1/2
d)5/8
e)3/4
I solved this question by assuming that order does not matter:
We have two cases in which the sum is odd: odd*odd*odd or even*odd*even
So it is 50/100*50*100*50/100 + 50/100*50/100*50/100. This yields Â¼ while the correct answer is Â½. After reading this thread, I applied combinations (order does NOT matter) and was able to get Â½
50c1 * (50c2 +50c1)/100c3
I have a few questions:
a) Can I solve this problem using permutation? If yes, can I just consider that I still have two cases: OOO + EOE and then apply permutation instead of combination when performing calculations or do I have to consider all the other possibilities first (OOO+EEO+EOE+OEE) and then perform math using permutation?
b) Why do I get the wrong answer with my simple calculation 50/100*50*100*50/100 + 50/100*50/100*50/100.
c) If order is assumed to matter when we use consecutive multiplication, then shouldn't have I gotten the wrong answer to the question above when I applied combinations?
I thank you in advance for the valuable time you'll spend out of your busy schedule on this and apologize for the long post.

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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 am; Thanked: 2256 times; Followed by:1535 members; GMAT Score:800

by lunarpower » Fri Jul 30, 2010 5:06 am

tnaim wrote:If I understand you correctly, did you say in the posts above that we can pretty much solve any probability problem either way: by considering that order matters or by considering that order does NOT matter as long as we are consistent in applying our pick to both the number of favorable outcomes and the number of possible outcomes?

that's partially true.
it's only completely true, though, if you also include the solution technique itself (i.e., the actual arithmetic and manipulations that you use to solve the problem) in your analysis.
that matters, a lot, because many solution techniques only work for one or the other of these 2 cases (order matters vs. order doesn't matter).
for instance, the most important case that comes to mind is the multiplication of consecutive probabilities. if you multiply consecutive probabilities, then order ALWAYS matters; it's actually impossible to use consecutive multiplication if you formulate the problem in such a way that order doesn't matter.

did I get that right? if I have, I think that the implications are very important and one of them is: I no longer have to think whether order matters in any probability problem.

... so, no, that's not true (unfortunately), for the reason mentioned above.

you still have to consider whether order matters, so that you can judge whether a particular technique of solution will work.

I can simply learn how to solve any probability problem by considering that order does NOT matter and just make sure I apply that to both the number of favourable and possible outcomes

well, no.

first, this approach would fail in any case in which order explicitly matters.
for instance, if i ask explicitly for the probability of getting even, THEN even, THEN odd, you won't be able to do the problem without accounting for that order.
also, in cases *with* replacement -- such as the problem that's in this thread -- you also can't use "order doesn't matter" approaches (combinations), since the combination formula REQUIRES that the choices be made without replacement.

second, even in cases when this statement is nominally true, it's sometimes still a bad idea -- there are lots of problems that can be solved with either approach, but are much more easily solved with the "order matters" approach.
for instance, if you want the probability of picking 2 black cards in a row from a 52-card deck, it's much easier to calculate (1/2) x (25/51) than it is to bother with (26c2)/(52c2).

specific responses for this problem below.

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by lunarpower » Fri Jul 30, 2010 5:07 am

I solved this question by assuming that order does not matter:
We have two cases in which the sum is odd: odd*odd*odd or even*odd*even

this approach doesn't work because, as stated above, you're using consecutive multiplication -- in which case order MUST matter. you can't multiply probabilities like this if order doesn't matter.

so, if you're going to do this, you must assume four possibilities:
odd, odd, odd
even, even, odd
even, odd, even
odd, even, even

So it is 50/100*50/100*50/100 + 50/100*50/100*50/100.

add two more 50/100*50/100*50/100's, for the reasons mentioned above. if you do that, you'll get the right answer.

This yields Â¼ while the correct answer is Â½. After reading this thread, I applied combinations (order does NOT matter) and was able to get Â½
50c1 * (50c2 +50c1)/100c3

i don't have pencil, paper, or a calculator, so i can't work this out right now, but i can definitely tell you that this expression does not equal 1/2. (i can tell this without actually evaluating the expression: note that 100c3 = (100x99x98)/(3x2x1) will contain the prime factor 11, and that won't get canceled out anywhere else in the expression.)
how did you figure this was 1/2?

also, another reason you can definitely tell this is wrong is that it contains 100c3, an expression that is inappropriate in this problem.
100c3 refers to the choice of three objects without replacement from a set of 100 objects. since this problem involves choosing objects with replacement, any approach containing the expression 100c3 is pretty much automatically going to be misguided.

I have a few questions:
a) Can I solve this problem using permutation? If yes, can I just consider that I still have two cases: OOO + EOE and then apply permutation instead of combination when performing calculations or do I have to consider all the other possibilities first (OOO+EEO+EOE+OEE) and then perform math using permutation?

you can't use "permutation formulas" at all, actually. those also require choices WITHOUT replacement.

by the way, "permutation formulas" are pretty much always a waste of time, since they can be replaced with simple consecutive multiplication. for instance, if i want a permutation of three objects from a pool of 15 objects, it's pretty silly to write 15!/(15-3)! and reduce it, when i can just use consecutive multiplication and write 15x14x13.
(the only real use of "permutation formulas" is in writing general expressions, in which "n" and "r" are variable expressions -- something you will absolutely never have to do on this test. if you only work combinatorics problems with hard numbers, as will be the case on this exam, then consecutive multiplication is always better and faster than "permutation formulas".)

"combination formulas", on the other hand, do have some degree of usefulness.

b) Why do I get the wrong answer with my simple calculation 50/100*50*100*50/100 + 50/100*50/100*50/100.

see above -- you're only considering half the possibilities.

c) If order is assumed to matter when we use consecutive multiplication, then shouldn't have I gotten the wrong answer to the question above when I applied combinations?

actually, you did get the wrong answer.
again, i don't know what that fraction actually is, but it's not 1/2.

Last edited by lunarpower on Fri Jul 30, 2010 5:52 pm, edited 1 time in total.

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by Ian Stewart » Fri Jul 30, 2010 2:56 pm

I agree with what Ron says above, and I'd just add a couple of things:

* When I said in another post that you'll get the right answer whether you consider order to matter or not, I was referring to a specific (but very common) type of problem - one in which you are selecting two or more objects without replacement from a set. So if you have 20 socks in a drawer, 8 of which are red, and you are asked the probability of picking two red socks if you take two socks out of the drawer (without replacement), you can look at this in one of two ways, either of which is perfectly correct: you can imagine picking the socks one at a time (i.e., in order), or you can imagine picking the two socks simultaneously (i.e. where order does not matter). If we imagine order to matter, the probability the first sock is red is 8/20, and the probability the second is red is 7/19; multiplying these, we get the answer. If we imagine order not to matter, there are 8C2 ways of picking two red socks, and 20C2 ways of choosing any two socks, so we get 8C2/20C2 as the answer. These two answers are equal, of course. In simple probability problems, I usually find it easiest to imagine order matters, at least when that approach is available. In more complicated questions (which don't tend to appear on the GMAT, though I often see them in prep books), assuming order does not matter is sometimes the only practical approach (here I'm thinking of the kinds of problems you see in an undergraduate combinatorics class - for example, if you are asked to find the probability of being dealt a full house in poker - but you don't see problems that difficult on the GMAT).

* Ron makes a very good point about the permutation formula, and I have no idea why it appears in so many prep books. Mathematicians need that formula to prove general results about counting, but you aren't preparing to do an undergraduate degree in combinatorics - you're preparing for the GMAT. The permutation formula is never required on the test. Sure, if you are asked how many ways you can choose a President and Vice-President from a group of 8 people, you'll get the right answer by calculating 8!/(8-2)!, but it's more direct to simply multiply your choices for each position: 8*7.

* The problem you post above, about the numbered balls, can actually be solved quite quickly. After we pick two of the balls, the sum is either even or odd. If the sum is even, we need to pick an odd ball - the probability of this is 1/2. If the sum is odd, we need to pick an even ball - the probability of this is 1/2. So no matter what has happened after we make our first two selections, there will be a 1/2 probability that our third selection will give us an odd sum, and the answer must therefore be 1/2.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Fri Jul 30, 2010 5:44 pm

So many experts chiming in on combinations!

Here's why we all love these types of questions: they're ideally suited to critical thinkers.

In my mind, one of the most important take-aways in this thread is what Ian said:

Sure, if you are asked how many ways you can choose a President and Vice-President from a group of 8 people, you'll get the right answer by calculating 8!/(8-2)!, but it's more direct to simply multiply your choices for each position: 8*7.

In general, test takers rely way too much on formulaic approaches to counting and probability problems on the GMAT. Of all of the online Kaplan resources, the toughest math quiz is the Permutations and Combinations Quiz. Yes, every question in that quiz can be solved via formulae. However, in almost every case, a critical thinking/strategy based approach yields a much quicker solution.

Remember, the GMAT is not a test of whether you can memorize a bunch of formulae. Sure, knowing the basic math underlying the quant section is a must, but the way you get a fantastic quant score is by mastering the critical thinking and strategies applicable to the GMAT.

Accordingly, every time you review a math question you do (counting or otherwise), you should ask yourself: "was there another approach I could have taken to get this answer faster and/or more confidently?" Once you've started thinking along those lines, you'll find that when you write CATs you're far more likely to hit on the best approach on each question.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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tnaim: Senior | Next Rank: 100 Posts; Posts: 67; Joined: Tue Mar 16, 2010 2:06 pm; Thanked: 3 times

by tnaim » Sun Aug 01, 2010 12:52 pm

Ron, Ian, and Stuart
Thank you all for contributing; how noble it is of each one of you to spend your own valuable time sharing with us what you know. You have reminded me of how powerful sharing can be.
Ron and Ian,
Spot-on answers. I now know that I can choose whether order matters only in probability questions that do not have "replacement" and in which we are selecting two or more objects. In addition, I have a clearer sense now as to why "permutation" is rather a waste of time for GMAT purposes. Another rule of thumb that I have learnt is that order DOES matter when calculating consecutive multiplication probabilities. The only question that I have remaining in this regard is this:
Ron, I mistyped the calculation of the combination solutions I wrote for the question above. Here is the solution that yields Â½
OOO or EOE
OOO (50!/(50-3)!3!)/100!/(100-3)!3! . result is 4/33
EOE (50!/(50-2)!2!*50)/100!(100-3)!3!. result is 25/66
The answer is 33/66.
I just wanted to confirm that the fact that I got Â½ using combinations (which is incorrect to use in this case) is just coincidental.
Stuart,
Thank you for the valuable feedback; I'm realizing how important it is to not only get the right answer within say 3 minutes but to also see what other methods are out there that can help shorten the time it takes to solve a question. A case in point is the tutorial that MGMAT on fast math that has helped me cut down the time I spend on such problems
https://www.manhattangmat.com/tutorials/ ... -score.cfm
unfortunately though, you would have to spend a lot of time searching for these alternative solutions and identifying which ones work for you, and it's REALLY hard to push yourself to do that on questions for which you get correct answer within the 2-2:30 minutes mark.

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by lunarpower » Mon Aug 02, 2010 1:52 am

tnaim -- thanks for the kind words regarding the previous posts.

tnaim wrote:The only question that I have remaining in this regard is this:
Ron, I mistyped the calculation of the combination solutions I wrote for the question above. Here is the solution that yields Â½
OOO or EOE
OOO (50!/(50-3)!3!)/100!/(100-3)!3! . result is 4/33
EOE (50!/(50-2)!2!*50)/100!(100-3)!3!. result is 25/66
The answer is 33/66.
I just wanted to confirm that the fact that I got Â½ using combinations (which is incorrect to use in this case) is just coincidental.

yes, it's coincidental; you are actually solving a different problem.

the problem that you are solving here involves selecting three balls without replacement.
intuitively, the reason why you still get an overall probability of 1/2 is because the two cases that you require (odd, odd, odd; even, odd, even) are symmetrical to the other two cases, which you don't want. ("symmetrical" in the sense that they have the same chance of occurring -- remember that there are as many odd as even balls in this situation.) therefore, in this situation, if you take any specific outcome and switch all the "E"s and "O"s, you must get another outcome with exactly the same probability as the original one.
specifically, OOO is symmetrical with EEE, and EEO is symmetrical with OOE (note the switching of O's and E's). therefore, the aggregate probability of the two possibilities that you do want (OOO, EEO) must be the same as the aggregate probability of the two possibilities that you don't want (EEE and OOE). since those two groups of possibilities together account for everything that can possibly happen, each of them must have an aggregate probability of 50%.

try the problem both ways in a scenario when it is not symmetrical; for instance, consider the selection of only two balls.
* in this case, the probability of getting an odd result with replacement is still 50%; you can get this result either from raw calculation (EO + OE = (1/2)(1/2) + (1/2)(1/2)) or by using ian's rather brilliant method above, in which you realize that, once you've picked the first ball, you then need either an even ball (if the first ball is odd) or an odd ball (if the first ball is even); in either case, the chance of getting the type of ball that you actually want is 1/2.
* on the other hand, without replacement, the chance of getting an odd result is now more than 1/2. here's the calculation with consecutive multiplication: OE + EO = (1/2)(50/99) + (1/2)(50/99) = 50/99. if you use your method of combinations (which is now allowed, since we are working the problem without replacement), you should arrive at the same result.
note that the symmetry argument doesn't work anymore, since the two possibilities that give you an even result (EE and OO) are now symmetric with each other, not with the two possibilities that give you an odd result (OE and EO, which, interestingly, are also now symmetric with each other).

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

by sanju09 » Mon Aug 02, 2010 3:20 am

It has been very educational thread, really. Thanks BTG for having such great caring and sharing instructors like Ron and the Stuart couples.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

by Stuart@KaplanGMAT » Mon Aug 02, 2010 7:06 pm

sanju09 wrote:It has been very educational thread, really. Thanks BTG for having such great caring and sharing instructors like Ron and the Stuart couples.

Hey - we went out for a beer once, I'm not sure we're a couple!

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 am; Location: India; Thanked: 267 times; Followed by:80 members; GMAT Score:760

by sanju09 » Tue Aug 03, 2010 6:10 am

Stuart Kovinsky wrote:
sanju09 wrote:It has been very educational thread, really. Thanks BTG for having such great caring and sharing instructors like Ron and the Stuart couples.
Hey - we went out for a beer once, I'm not sure we're a couple!

excess of beer costs deer sometimes

never mind, it has been the greatest joke of the day. Excuse me for driving so closely in front of you. Anything worth taking seriously is worth making fun of. Better to understand a little than to misunderstand a lot. Let me reiterate...

It has been very educational thread, really. Thanks BTG for having such great caring and sharing instructors like Stuart Kovinsky, Ian Stewart, and Ron, who sometimes cannot bear beer

cheer

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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tnaim: Senior | Next Rank: 100 Posts; Posts: 67; Joined: Tue Mar 16, 2010 2:06 pm; Thanked: 3 times

by tnaim » Sun Dec 05, 2010 1:28 pm

Experts!
The issue of whether order matters is confusing me again.
Ron,
from what I understood in your previous posts, you said that it's impossible to solve consecutive probability questions without considering that order matters. I just wanted to clarify whether consecutive probability means "choosing one item at a time, then choosing a next one right after it, with replacement"? The reason why I got confused is the question below that I got on a practice exam:

MGMAT CAT question

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y â‰¤ 8
(2) y â‰¥ 4

the answer to this question is B. B is sufficient on the ground that you can have RW or WR (red, white) or (white, red).
I could not understand why order would matter here. We don't necessarily have the following two conditions:
1) no replacement: the question does not mention that the balls are returned back
2) I can take out two marbles at the same time -- grab two at a time and not one after another. in that case, why would order matter? as long as I have two balls, each of which is of a different color, then that's all that matters.

Ian,
I think that the example above is similar to the example that you mentioned in your post -- choosing 2 socks from 8.

Stuart,
I know that you have solved this MGMAT CAT question using trial and error on another thread, but I did not feel comfortable going that route during the exam as I was afraid I won't be able to cover all possibilities on time.

many thanks to all of you!!!

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aditya_velma: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Sep 09, 2010 6:00 am

by aditya_velma » Tue Dec 07, 2010 12:03 pm

Hey tnaim,
I have been following this thread to learn more about the 'order' in probability, but i guess I can help you with this particular prob. Lets try:

Q: is it more likely that she will have 2 red marbles than that she will have one marble of each color?

Condition for YES: prob(2R)> prob(1R,1W)

P(2R)= 8C2/(8+y)C2 = 8*7/[(8+y)(8+y-1)]

P(1R,1W) = (8C1*yC1)/[(8+y)C2] = [2*8y]/[(8+y)(8+y-1)] note that we have used combitronics here so: "order is not assumed" (the 2 in 2*8y comes from the combination in denominator)

if we use multiplication:"order is assumed by default" --->
P(1R,1W) = P(1R1W) + P(1W1R)= [8/(8+y)] [y/(8+y-1)] + [y/(8+y)] [8/(8+y-1)] = [2*8y]/[(8+y)(8+y-1)]

'Here u can see that in both approaches P(1R,1W) resulted in same expression'

now lets substitute P(2R) and P(1R,1W) in the condition for YES:
prob(2R)> prob(1R,1W)
(8*7)>(2*8y) as denominators [(8+y)(8+y-1)] get canceled on both sides
3.5>y
y<3.5 for a YES

Statement 2 says y=or> 4, in which case the condition will be a definite NO. Hence, B sufficient

Ron,
I have a question here. I understood when u said that certain problems may require order etc.
In the problem https://www.beatthegmat.com/combinatorics-t71136.html

when I say (90/800)*(1/600), why shouldn't I add (90/600)*(1/800) to this? Is it because we are selecting 1 piece each from 800 and 600 respectively, instead of 2 pieces from (800+600) for the total number of conditions?

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The confusion over order in probability...

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