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10 Oct 2005, 05:36
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B
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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%
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%
10 Oct 2005, 06:21
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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.
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.
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26 Nov 2014, 11:39
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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
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
