Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong.
From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.
Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
10 Oct 2005, 04:36
25
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
25% (02:40) correct 75% (03:02) wrong based on 265 sessions
HideShow timer Statistics
Darcy, Gina, Ray and Susan will be the only participants at a meeting. There will be three soft chairs in the room where the meeting will be held and one hard chair. No one can bring more chairs into the room. Darcy and Ray will arrive simultaneously, but Gina and Susan will arrive individually. The probability that Gina will arrive first is 1/3, and the probability that Susan will arrive first is 1/3. The probability that Gina will arrive last is 1/3, and the probability that Susan will arrive last is 1/3. Upon arriving at the meeting, each of the participants will select a soft chair, if one is available.
If Darcy and Ray arrive and see only one unoccupied soft chair, they will flip a fair coin to determine who will sit in that chair. By what percent is the probability that Darcy will sit in a soft chair greater than the probability that Gina will sit in a soft chair?
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
10 Oct 2005, 05:21
5
3
I got B. This is how I got it. The prob that Darcy will sit in a soft chair=prob(arriving first)+prob(arriving second)+(0.5)prob(arriving third). If she arrives third, she will be arriving with Ray and since there will be one soft chair remaining they flip a coin for it. That is why I halfed the last term in the prob. Since we know that the prob of Gina or Susan to arrive first is 1/3 each, then the prob of Darcy and Ray to arrive first is also 1/3. From this knowledge we, can use the info in the stemn to deduce that the prob of Darcy and Ray to arrive second or last is also 1/3 for each instance. Therefore
prob of getting soft chair is 1/3+1/3+(0.5)(1/3)=5/6
The prob of Gina sitting in the soft chair, can happen only if she arrives first or second, so the prob=1-prob(arriving last)=1-1/3=2/3.
From here, use [(5/6)-(2/3)]/2/3
This equals 25% (b). I know this was very confusing, and I may not even be correct. What is the OA? If I am correct, I will try an explain it more if needed.
_________________
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
10 Oct 2005, 05:25
These are all the possible combinations:
S takes a hard chair
DGRS--1/6
GRDS--1/6
D takes a hard chair
SGRD--1/6
GRSD--1/6
R takes a hard chair
SDGR--1/12
DSGR--1/12
G takes a hard chair
SDRG--1/12
DSRG--1/12
Therefore, the probability that Darcy will sit in a hard chair is 1/3 and the probability that Darcy will sit in a soft chair is 2/3
And the probability that Gina will sit in a hard chair is 1/6 and the probability that Gina will sit in a soft chair is 5/6.
By what percent is the probability that Darcy will sit in a soft chair greater than the probability that Gina will sit in a soft chair? . This is a ticky question to answer as the first probablity is lower than the second one.
I would answer 0.
If the question were asking the opposite, ie by what percent the probability that Gina will sit in a soft chair greater than the probability that dary will sit in a soft chair is, the answer would be 25%.
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
10 Oct 2005, 07:55
GMATT73 wrote:
The prob of Gina sitting in the soft chair, can happen only if she arrives first or second, so the prob=1-prob(arriving last)=1-1/3=2/3.
From here, use [(5/6)-(2/3)]/2/3
These were the two crucial steps that stumped me. Thanks for putting it all together Chris
The probablity that Gina arrives third and does not get a soft chair really depends on whether Darcy and Ray want to take up a soft chair at all Doesnt it?
If Susan comes first, and picks the hard chair, then it doesn't really matter if Gina comes third or even last since she will always gets soft chair in this instance. To me, this question has a lot of assumptions.
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
10 Oct 2005, 09:10
I assume that if there is an empty soft chair, you must take it. The stem states, "Upon arriving at the meeting, each of the participants will select a soft chair, if one is available".
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
12 Oct 2005, 17:05
chriswil2005 wrote:
I got B. This is how I got it. The prob that Darcy will sit in a soft chair=prob(arriving first)+prob(arriving second)+(0.5)prob(arriving third). If she arrives third, she will be arriving with Ray and since there will be one soft chair remaining they flip a coin for it. That is why I halfed the last term in the prob. Since we know that the prob of Gina or Susan to arrive first is 1/3 each, then the prob of Darcy and Ray to arrive first is also 1/3. From this knowledge we, can use the info in the stemn to deduce that the prob of Darcy and Ray to arrive second or last is also 1/3 for each instance. Therefore prob of getting soft chair is 1/3+1/3+(0.5)(1/3)=5/6
The prob of Gina sitting in the soft chair, can happen only if she arrives first or second, so the prob=1-prob(arriving last)=1-1/3=2/3.
From here, use [(5/6)-(2/3)]/2/3
This equals 25% (b). I know this was very confusing, and I may not even be correct. What is the OA? If I am correct, I will try an explain it more if needed.
I believe that your last step should just be 1/6*100% = 16.66%. We're asking percentage differecne and since we already found the difference (by subtracting the probability), then we should just conver it to a percentage change.
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
26 Nov 2014, 10:39
1
2
We will do:
Darcy = D Ray = R Susan = S Gina = G
Keep in mind that follows:
1 Darcy and Ray arrive together at the location of the chairs.
2. the probability that Susan arrives first is 1/3 and the probability that Gina arrives first is also 1/3.
3. the probability that Gina comes last is 1/3 and the probability that Susan comes last also is 1/3.
1.-deduce that DR, come as a unit to the location of the chairs, so we will have three candidates (not four) arriving at the location of the chairs, these are: DR, S and G.
2.-given that there are three candidates for the position of the chairs, it follows that the probability of arriving first DR is 1/3.
3.-given that there are three candidates for the position of the chairs, it follows that the probability of arriving last DR is 1/3
We can also deduce from all of the above information to: as each of the three: DR, G or S, may become second and that between the probability that arrive in the first or the last place each of the three add 2/3, we conclude that each of the three has a probability of 1/3 in second place.
Thus we have arrival location of chairs
DR probability:
For 1st place = 1/3 For 2nd place = 1/3 For 3rd place = 1/3
Probability of S:
For 1st place = 1/3 For 2nd place = 1/3 For 3rd place = 1/3
Probability of G:
For 1st place = 1/3 For 2nd place = 1/3 For 3rd place = 1/3
So:
Location of D to access a soft Chair
If DR comes in first or second place both as R D they will access a soft Chair, this means that D in this situation has 1/3 + 1/3 to a soft Chair.
If DR comes third (last), D has a probability of (1/2) x (1/3) access to a soft Chair.
Then in all cases D has a probability of 1/3 + 1/3 + (1/2) x (1/3) = 5/6 access to a soft Chair.
Location of G to access a soft Chair
To G enter a soft Chair, you can only get in first or second place, if it reaches third we have addressed three soft chairs DR two soft chairs and S a soft Chair. The probability that G arrives in first or second place is 1/3 + 1/3 = 2/3 = 4/6
Finally the highest percentage who D on G accesses a soft Chair is:
100 x (5/6 - 4/6) /(4/6) = 25% correct response b)
hurtado Claudio gmat gre sat math preparation
_________________
claudio hurtado maturana Private lessons GMAT QUANT GRE QUANT SAT QUANT Classes group of 6 students GMAT QUANT GRE QUANT SAT QUANT Distance learning courses GMAT QUANT GRE QUANT SAT QUANT
Website http://www.gmatchile.cl Whatsapp +56999410328 Email clasesgmatchile@gmail.com Skype: clasesgmatchile@gmail.com Address Avenida Hernando de Aguirre 128 Of 904, Tobalaba Metro Station, Santiago Chile.
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
09 Oct 2016, 03:42
Hello Guys I Understood the process of getting 2/3 and 4/5 . But I do not understand why we divided the difference between them by 2/3. Could someone please explain that for me?
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
09 Oct 2016, 06:10
nawaf52 wrote:
Hello Guys I Understood the process of getting 2/3 and 4/5 . But I do not understand why we divided the difference between them by 2/3. Could someone please explain that for me?
Because we have to find how much higher is D's probability (of getting a soft chair) than G's one. Not the absolute difference between the two probablities. Is it clear ?
_________________
03/29/16 GmatPrep CAT1 600 (Q 47 V 26) + long break. 09/26/16 MGMAT CAT1 560 (Q41 V 27) 10/12/16 MGMAT CAT2 630 (Q46 V 31) 10/21/16 MGMAT CAT3 640 (Q42 V36) 10/28/16 GmatPrep Cat2 660 (Q49 V 31) 11/15/16 MGMAT Cat4 640 (Q41 V 36)
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
16 Jul 2017, 21:48
GMATT73 wrote:
Darcy, Gina, Ray and Susan will be the only participants at a meeting. There will be three soft chairs in the room where the meeting will be held and one hard chair. No one can bring more chairs into the room. Darcy and Ray will arrive simultaneously, but Gina and Susan will arrive individually. The probability that Gina will arrive first is 1/3, and the probability that Susan will arrive first is 1/3. The probability that Gina will arrive last is 1/3, and the probability that Susan will arrive last is 1/3. Upon arriving at the meeting, each of the participants will select a soft chair, if one is available.
If Darcy and Ray arrive and see only one unoccupied soft chair, they will flip a fair coin to determine who will sit in that chair. By what percent is the probability that Darcy will sit in a soft chair greater than the probability that Gina will sit in a soft chair?
Experts, I need help with some questions regarding this.
1.Are we keeping track of the order in which they arrive AND take the chairs? or just the order in which they arrive?
2.How do we create the complete sample space for this question?
3.What is the probability of the event - Gina arrives first, Susan arrives second, and Darcy and Ray arrive last?
4. The probability that some one will arrive first, somebody second, and somebody third are all individually 1 (1/3 + 1/3 + 1/3). The probability that all of them will arrive is also 1, as we are sure all of them will certainly arrive at some point. From the provided information, how do I calculate this probability? in other words, how to make the probabilities sum up to 1?
This will solve many problems for me and make me much better at probability. Please help.
_________________
Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
[#permalink]
Show Tags
10 Aug 2018, 09:26
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
close
25% (02:40) correct 
