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Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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10 Oct 2005, 05:36
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26% (02:51) correct 74% (03:06) wrong based on 203 sessions
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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? A. 50% B. 25% C. 16 2/3% D. 12 1/2% E. 0%
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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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10 Oct 2005, 06:21
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=1prob(arriving last)=11/3=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.




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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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10 Oct 2005, 06:25
These are all the possible combinations:
S takes a hard chair
DGRS1/6
GRDS1/6
D takes a hard chair
SGRD1/6
GRSD1/6
R takes a hard chair
SDGR1/12
DSGR1/12
G takes a hard chair
SDRG1/12
DSRG1/12
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%.



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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10 Oct 2005, 06:36
The prob of Gina sitting in the soft chair, can happen only if she arrives first or second, so the prob=1prob(arriving last)=11/3=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



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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10 Oct 2005, 08:55
GMATT73 wrote: The prob of Gina sitting in the soft chair, can happen only if she arrives first or second, so the prob=1prob(arriving last)=11/3=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.



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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10 Oct 2005, 10:10
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".
GMATT, what is the OA?



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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12 Oct 2005, 18:05
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=1prob(arriving last)=11/3=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.



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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13 Oct 2005, 08:44
OA is B. This is from the Kaplan course book. OE is a full page in length, but Chris pull it together in a nutshell.



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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26 Nov 2014, 11:39
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
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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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09 Oct 2016, 04:42
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?



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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09 Oct 2016, 07:10
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 ?
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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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16 Jul 2017, 22:48
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?
A. 50% B. 25% C. 16 2/3% D. 12 1/2% E. 0% IanStewart VeritasPrepKarishma BunuelExperts, 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.



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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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Re: Darcy, Gina, Ray and Susan will be the only participants at a meeting.
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