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Kaplan challenge problem!

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Kaplan challenge problem!

by GmatKiss » Sun Oct 16, 2011 6:29 am
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%

Source: Kaplan GMAT
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by GmatKiss » Sun Oct 16, 2011 6:29 am
Explanations needed!!
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by shankar.ashwin » Sun Oct 16, 2011 8:22 am
We need to find P(G) and P(D)

P(G) = 1-1/3 = 2/3 (1 - P(G will arrive last)

P(D) = 1/3 * 1/2 = 1/6 (Prob of D&R coming last and D losing the toss)
P(D) = 5/6

So (5/6-2/3)/2/3 = 1/4

B IMO
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by moonraker » Sun Oct 16, 2011 10:26 am
take ray and darcy as 1 unit = RD
Gina = G and Susan = S

Now if Gina comes first and second, then only she can get a soft chair.
Now P(G) coming first and last = 1/3, hence P(G) coming second is also = 1/3 ( Note: Ray and Darcy are one unit and arrive together - so effectively 3 sequences are possible - first, second and third).

Hence probability of Gina getting a soft chair = 1/3 + 1/3 = 2/3 .......... (a) ( If she comes last she does not get a soft chair - only 1 hard chair is left)

Probability of Darcy getting a soft chair is different - she can get chairs in all three sequences)

For first P(D-1) = 1/3
For second entry = 1/3 ( here 2 chairs are left and 2 people vying for them - darcy and ray)
for third entry P(G-3) = 1/3 x 1/2 ( only 1 chair and toss is needed)

Hence P(G) = 1/3 + 1/3 +1/6.................(b)

For percentage increase :

[ P(D) - P(G) ] / P(G) = (1/6) / (2/3) = 1/4 = [spoiler]25%[/spoiler]

hence answer = B
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