Advanced Probability - Who Gets a Seat?

beeparoo

Not only is it long, but it is complicated. Yet, despite knowing the answer, I have been stumped more than once. I welcome other unique approaches from various members.
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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%

Source: Kaplan GMAT

wawatan · Posted: Sat Jun 28, 2008 7:36 pm Post subject:

wow long problem.

Ian Stewart

beeparoo · Posted: Sun Jun 29, 2008 9:15 pm Post subject:

Wow, Ian. Your confidance throughout your solution is unwavering. I hope to build that level of "but of course!" attitude on GMAT.

Anyway, I was unsure of my footing during the calculation of Darcy's probability and especially Gina's, thinking that I should treat two instances separately:

For calculation of Gina as second arrival, consider the possibility that
1) Susan enters first
or
2) Darcy/Gina enter first

Thoughtfully reviewing your response, I see, in retrospect, how they don't need to be quantified separately. But hindsight is 20/20 and clearly, I need to polish my probability skills.