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Probability - number of Paths

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Probability - number of Paths

by ronnie1985 » Wed May 23, 2012 8:27 am
Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school located on a corner, lies three blocks south and three blocks east of his home, also located on a corner. If Alicia is equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability that she will walk south for the first two blocks?

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by GMATGuruNY » Wed May 23, 2012 12:05 pm
ronnie1985 wrote:Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school located on a corner, lies three blocks south and three blocks east of his home, also located on a corner. If Alicia is equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability that she will walk south for the first two blocks?

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Alicia's trip involves 3 movements south (SSS) and 3 movements east (EEE). We want to know the probability that her first 2 movements are SS. This question is no different from the following:

A bag contains three marbles labeled S and three marbles labeled E. If two marbles are randomly selected from the bag, what is the probability that both are labeled S?

P(SS) = 3/6 * 2/5 = 1/5.
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by Anurag@Gurome » Wed May 23, 2012 10:08 pm
ronnie1985 wrote:Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school located on a corner, lies three blocks south and three blocks east of his home, also located on a corner. If Alicia is equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability that she will walk south for the first two blocks?

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To reach the school, Alicia should walk 3 times south and 3 times east: SSSEEE
Total number of routes to the school = 6!/(3!3!) = 20 (Number of permutations of 6 letters out of which 3 S's and 3 E's are identical)
Now, we wan to count all the routes which start with {SS}. So, {SS} is fixed and then there can be any combination of the rest 4 letters SEEE.
So, all possible routes, which start with {SS} = 4!/3!=4 (number of permutations of 4 letters out of which 3 E's).

So, required probability = 4/20 = [spoiler]1/5[/spoiler].
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