I can't help you with you Q. But i wanted to know how you solved this. What is the answer?
Thanks, Rohit.
If there are 4 letters and 4 covers. What is the probability that exactly 1 letter goes into its exact cover?
Set up the basic problem:
__(C1) __(C2) __(C3) __(C4)
If we ignore the uncertainty of which is the cover for a moment, we can first just consider the question as if we were putting the correct letter in the first cover.
This makes: __(C1) _x_(C2) _x_(C3) _x_(C4)
Now, put the numbers we know into the question.
1/4(C1) 3/4(C2) 2/4(C3) 1/4(C4)
So the probability of getting the correct letter in the first cover is 3/128.
Now let's consider the fact that it could have been any letter in any cover.
This is what I refer to as rearrangement. The option is the exact same, all that changes is the relative positions of the places. So use the rearrangement formula.
# things moving (# things of type 1)(# things of type 2)...
Here we have 4 things moving. Each is a different letter, which tells us the formula is:
4/(1)(1)(1)(1) = 4
Finally, take the probability of the first option and multiply it by the number of rearrangements.
Of these 6 choices the choices in which all the 3 letters are in worng covers are choice3 and choice5. So there are two ways of doing so.
So the total number of ways in which only 1 letter is placed in the right cover when we have 4 letters and 4 covers is = 2*4(the right letter cover combination could be any of the four) = 8
The total number of ways of putting the letters into covers is 4! = 24.
So, the required probablility = 8/24 = 1/3
The other day my roommate and I found a shortcut for this type of problems on the net.
The answer is simply given by 1/(n-1); where n is the number of letters/covers.
Check this when n=3. From the above 6 ways of putting the covers in the letters; there are 3 ways in which exactly 1 lertter is in its proper cover. So required probability is = 3/6 = 1/2. (Formula works)
Hey Hashim, I am afraid the answer you arrived at is wrong. Let me explain. ... The answer is simply given by 1/(n-1); where n is the number of letters/covers.
Check this when n=3. From the above 6 ways of putting the covers in the letters; there are 3 ways in which exactly 1 letter is in its proper cover. So required probability is = 3/6 = 1/2. (Formula works)
Regards, Adarsh.
Adarsh,
You are right. I completely ignored the the outcomes with the other letters and covers. That is what happens to me when I am up too late or too early without enough sleep. My apologies to anyone confused.
Some of the differences in our solutions come from the assumptions we each make in solving the problem, which is why the GMAT has to be very careful in how it phrases questions. I normally take great care to point out where I think examples are problematic and unlike the GMAT. I blame fatigue again for not doing that here.
You assume with your solution that the selection/placement is sequential. I assumed that the selection is simultaneous.
Note that adjusting for my oversight and either assumption this only requires a small change to the METHOD that I showed above. I try to always emphasize to students that learning the method is what gets you problems right on the GMAT, not getting any particular example correct.
If you change the initial scenario to assume sequential selection (decreasing denominators) and incorrect letters only in other covers (fewer choices for each position), you get:
1/4(C1) 2/3(C2) 1/2(C3) 1/1(C4) = 1/12
The number of rearrangements and how to find them does not change.
1/12 * 4 = 1/3
Same answer. ;-)
I teach students to REDUCE the number of different approaches, formulas and facts that they learn for the GMAT. The GMAT like any standardizedtest is pattern oriented and will reuse the same information in the same way over and over again. It is to the students benefit to have one consistent approach to a GROUP of problems far more so than to have a list of formulas that each work with one variation.
Hey Hashim, Thanks for the explanation. As said earlier, I was able to do this for 4 letters/covers, but did not know the general procedure for more than 4 letters/covers.
Your method helps doing this.
Moreover can you just explain me how to solve the following questions.
1. Probability of putting all letters in wrong covers. 2. Probability of putting only 1 letter in the right cover. (this is already explained). 3. Probability of putting 2 letters in the right cover and so on.
I guess there is a general formula for this. I remember doing this sometime in my Inter while preparing for Eamcet. Its been 9 years since, so cannot recall. Can you help me in doing so?
Moreover one general question. I understand that you are a GMAT trainer or so from the posts you have written. I have just gone through your posts in SC. I need some advise on GMAT test taking strategies. I understand this thread is not the right one for thesde type of questions. Please let me know where I can contact you?
This is a bit urgent as I have only 5 days before my test date. Please reply.
Hey Hashim, Thanks for the explanation. As said earlier, I was able to do this for 4 letters/covers, but did not know the general procedure for more than 4 letters/covers.
Your method helps doing this.
Moreover can you just explain me how to solve the following questions.
1. Probability of putting all letters in wrong covers. 2. Probability of putting only 1 letter in the right cover. (this is already explained). 3. Probability of putting 2 letters in the right cover and so on.
I guess there is a general formula for this. I remember doing this sometime in my Inter while preparing for Eamcet. Its been 9 years since, so cannot recall. Can you help me in doing so?
Moreover one general question. I understand that you are a GMAT trainer or so from the posts you have written. I have just gone through your posts in SC. I need some advise on GMAT test taking strategies. I understand this thread is not the right one for thesde type of questions. Please let me know where I can contact you?
This is a bit urgent as I have only 5 days before my test date. Please reply.
Regards, Adarsh.
Adarsh,
You probably should not obsess over strange probability questions, you will have at most 3 advanced stats questions, and at most 2 probability and that is RARE. Understand the pattern and process I laid out and you will be fine.
All of those questions can be solved using the same method I laid out in my previous post, with adjustments for the new information:
1. Probability of putting all letters in wrong covers.
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x y is: A. 65 B. 55 C. 45 D. 35 E. 25
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x - y is: A. 65 B. 55 C. 45 D. 35 E. 25
Interesting Question , from my favorite Venn Diagrams Answer is B. 55 In a Venn Diagram , let's say DVD player are A , Cell Phone are B . MP3 Players are C X ( maximum ) : Maximum no of Households have all A,B and C is 55 bcoz C cannot be more 55 . Y ( Minimum ) : Minimum no of Households have all A,B and C is 0 bcoz it can be distribute like 25 Households have A and C . 50 Households have A and C . 30 Houses have B and C . 5 Houses do not have anything , hence Leaving Households having all the three are 0 .
Hence X-Y = 55-0 =55 . Please correct me if anything is wrong
I too got the ans as 55 i.e B Total households having both DVD and cellph be a, cellph and mp3 be b and mp3 and dvd be c When x is max, a b and c are 0 hence x (max) = 55 For y (min), y = 0 hence x-y = 55
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x y is: A. 65 B. 55 C. 45 D. 35 E. 25
Ashish,
This is a very interesting question. I particularly like math questions that add nuances to common concepts (my first degree is in mathematics).
Unfortunately, this question probably goes beyond the scope of anything that you are likely to see on the GMAT. Questions that go this far outside of the GMAT style of writing questions have both good and bad points.
GOOD: The question can help you develop a confidence and competence that exceeds anything you need for the test.
BAD: The question encourages you to develop competence BEYOND WHAT YOU NEED FOR THE TEST. Not only that, but this sort of question also causes students to focus too much on topics that seem far more difficult than they actually are, costing those students time and causing anxiety in an already pressure filled and stressful situation.
There is already a good solution posted, so I will leave it at that.
I too got the ans as 55 i.e B Total households having both DVD and cellph be a, cellph and mp3 be b and mp3 and dvd be c When x is max, a b and c are 0 hence x (max) = 55 For y (min), y = 0 hence x-y = 55
wats the oa?
thanks
answer is 45 maximum X is 55 but minimum Y is 10 ....which makes the answer 45. here i understood that the answer csnnot be 55 caz making Y min to 0 will not sum up to 100.
This is a very interesting question. I particularly like math questions that add nuances to common concepts (my first degree is in mathematics).
Unfortunately, this question probably goes beyond the scope of anything that you are likely to see on the GMAT. Questions that go this far outside of the GMAT style of writing questions have both good and bad points.
GOOD: The question can help you develop a confidence and competence that exceeds anything you need for the test.
BAD: The question encourages you to develop competence BEYOND WHAT YOU NEED FOR THE TEST. Not only that, but this sort of question also causes students to focus too much on topics that seem far more difficult than they actually are, costing those students time and causing anxiety in an already pressure filled and stressful situation.
There is already a good solution posted, so I will leave it at that.
Hashim, This question is taken from GMAT practice test only. 800_score is the source for the question. so this is a GMAT type question.
Ashish ... search for commentary on the internet on their questions.
I have taught the GMAT since 1994 and I can guarantee you that you will NOT find a question like that on the actual test. I have taken the actual GMAT many times in both paper and computer incarnations. I have been teaching from official guides since version 9 (or maybe . Take my word for it, you do not want to spend a lot of time worrying about this particular question.
You may find on the GMAT a question that tests 3 Groups concepts (very rare but they do occur). You can certainly find questions that take a concept and add the maximization/minimization element to make it slightly harder. But, the combination of the two is something that would make any seasoned teacher skeptical. There are other things about the question that are also not very GMAT-like.
I am not telling you to stop using whatever resources you have to study with. But, I am suggesting that you should do so with the ability to make informed decision that few testtakers will actually have since you spend relatively little time with the test.
Ashish ... search for commentary on the internet on their questions.
I have taught the GMAT since 1994 and I can guarantee you that you will NOT find a question like that on the actual test. I have taken the actual GMAT many times in both paper and computer incarnations. I have been teaching from official guides since version 9 (or maybe . Take my word for it, you do not want to spend a lot of time worrying about this particular question.
You may find on the GMAT a question that tests 3 Groups concepts (very rare but they do occur). You can certainly find questions that take a concept and add the maximization/minimization element to make it slightly harder. But, the combination of the two is something that would make any seasoned teacher skeptical. There are other things about the question that are also not very GMAT-like.
I am not telling you to stop using whatever resources you have to study with. But, I am suggesting that you should do so with the ability to make informed decision that few testtakers will actually have since you spend relatively little time with the test.
answer is 45 maximum X is 55 but minimum Y is 10 ....which makes the answer 45. here i understood that the answer csnnot be 55 caz making Y min to 0 will not sum up to 100.
cud any1 explain how min y = 10 wat does da solution say?
I am having trouble learning the permutations and combinations,probability.Its so confusing and frustrating . Can you people give some pointers in how to prepare these topics thoroughly..What kind of materials are really useful in this regard.
I am having trouble learning the permutations and combinations,probability.Its so confusing and frustrating . Can you people give some pointers in how to prepare these topics thoroughly..What kind of materials are really useful in this regard.
Thanks in advance
RS,
Sorry to hear about your frustration.
In my experience, lots of people make the stats topics on the GMAT far more complicated than those topics warrant. There are only a few forms of the questions and a few variations that the test can make.
Most good books give simple clear explanations that are flexible enough to apply to make different types of questions. Some books/material does make the mistake of giving questions that are far more complex than those found on the GMAT.
You can feel free to check out the explanatory materials found at gmat.bellcurves.com (you will have to create an account but it is free). Under the prepare tab at the top of the page, go to the strategy pages. There you can find explanations and examples of some of the Stats topics. Hopefully, that will provide you with enough of a starting point that you instead are asking about what variations exist in particular questions that you might post here.
can any bdy plz solve Q206 OG 11 Page 180 problem solving the inscribed circle question ?? Could not Even understand the solution provided at the back of the section :confused:
Hey buddy, In the question, we divide the semicircle (upper half), into three arcs, arc OP, PQ and QR now, the sum of the angles subtended at the centre by all arcs constituting the semicircle should be 1800 .
Measure of arc PO: If an arc subtends an angle x at the circle, it will subtend an angle 2x at the centre. The angle subtended by arc PO at circle is 35(angle PRO), so it will subtend an angle of 70 at centre.
Similarly, the arc QR subtends and angle of 35 at the circle (Angle RPQ = 35 bcoz of alternate angles of parallel lines). Therefore, it subtends an angle 70 at the centre. Now, the only remaining arc is PQ, which will subtend (180-70-70=40 degree) at the centre. Now that we know the angle at the centre subtended by arc PQ, the length is (40/360)*2*PI*radius . Voila, u get the answer as 2*PI hth Sushant Bahadur
. Take my word for it, you do not want to spend a lot of time worrying about this particular question. 