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probability

by sacx » Thu Jan 22, 2009 2:36 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. 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. 12.5%
E. 0%
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by Ian Stewart » Thu Jan 22, 2009 3:20 am
Considering that they're describing a fairly simple situation, there are far too many words in the question! Anyway, I posted a solution here:

www.beatthegmat.com/advanced-probabilit ... 13122.html
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by gaggleofgirls » Thu Jan 22, 2009 8:56 am
To answer, we need to determine, the % that Gina gets a soft chair and the % that Darcy gets a soft chair.

If Gina arrives 1st or middle, she gets a soft chair, if she arrives last, she does not. 2 of the 3 times she can arrive, she gets the soft chair, so it is 2/3.

Darcy is very similar, if he arrives first, he (and Ray) arrive first, Darcy gets a soft chair, if he (and Ray) arrive middle, Darcy gets a soft chair, the only difference is that if he arrives last, he does not automatically get the hard chair like Gine, instead he has a 50% chance of getting the soft chair. 50% of 1/3 = 1/6 so Darcy's chance of getting the soft chair is 1/3+1/3+1/6 = 5/6

Now, what we need to know if the % difference between 5/6 and 2/3. I like to use a common denominator for this so it is the % diff between 5/6 and 4/6, which is what % greater is 5 than 4, so 25%.

Answer is B

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by krisraam » Thu Jan 22, 2009 12:10 pm
Answer id B -25%

My Approach is how many times does darcy or Gina get to sit on a Hard Chair.

First take Darcy

Darcy will sit in the Hard Chair only when Gina and Susan are first ans second( vice versa ) and Darcy has to lose the toss.

So the probabilty is :: 1/3(Gina Coming first) * 1/3( Susan coming second 1- (1/3(coming last) + 1/3(Coming first)) ) * 1/2 ( losing the toss to Ray) = 1/9 * 1/2

The same thing happens when Susan comes first

So 2* 1/9 * 1/2 = 1/9

The probablity Darcy gets to sit on a soft chair = 8/9

Now take Gina

Gina will sit on Hard chair only when Gina is the last person = 1/3

The probablity Gina gets to sit on softchair = 2/3

So % difference is (8/9-2/3)/(8/9) = 1/4 = 25%






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. 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. 12.5%
E. 0%
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