Probability

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Tue Oct 20, 2009 5:10 pm

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?

OA is
[spoiler]1/5[/spoiler]

Abdulla

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Source: Beat The GMAT — Problem Solving | Tue Oct 20, 2009 5:10 pm

uttam.albela: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Fri Aug 28, 2009 2:40 am; Location: Singapore; Thanked: 11 times

by uttam.albela » Thu Oct 22, 2009 6:25 am

Hi Abdulla,

a = To reach the destination, first find out total number of ways.
b = Then find out the number of ways in a particular manner mentioned in the problem.

Probability = b/a problem solved

Not let us find a and b.

First a:

Alicia has to cross 3 south and 3 east blocks. And, she can do this in any order.

So we need to find in how many ways we can arrange 3 south and 3 east blocks.

a = 6 ! / (3! * 3!) = 20

Now find b:

She has already traveled 2 south block, so she needs to travel 1 south and 3 east block.

b= 4 ! / (1! * 3 !) = 4

probability = b/a = 4/20 = 1/5

Let me know if anything is not clear to you.

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Thu Oct 22, 2009 6:47 pm

uttam.albela wrote:Hi Abdulla,

a = To reach the destination, first find out total number of ways.
b = Then find out the number of ways in a particular manner mentioned in the problem.

Probability = b/a problem solved

Not let us find a and b.

First a:

Alicia has to cross 3 south and 3 east blocks. And, she can do this in any order.

So we need to find in how many ways we can arrange 3 south and 3 east blocks.

a = 6 ! / (3! * 3!) = 20

Now find b:

She has already traveled 2 south block, so she needs to travel 1 south and 3 east block.

b= 4 ! / (1! * 3 !) = 4

probability = b/a = 4/20 = 1/5

Let me know if anything is not clear to you.

Thank you uttam.albela, it's clear now.

Abdulla

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ssuarezo: Master | Next Rank: 500 Posts; Posts: 126; Joined: Wed Jun 24, 2009 1:12 pm; Location: Montreal; Thanked: 2 times; GMAT Score:510

by ssuarezo » Thu Oct 22, 2009 8:32 pm

uttam.albela wrote:Hi Abdulla,

a = To reach the destination, first find out total number of ways.
b = Then find out the number of ways in a particular manner mentioned in the problem.

Probability = b/a problem solved

Not let us find a and b.

First a:

Alicia has to cross 3 south and 3 east blocks. And, she can do this in any order.

So we need to find in how many ways we can arrange 3 south and 3 east blocks.

a = 6 ! / (3! * 3!) = 20

Now find b:

She has already traveled 2 south block, so she needs to travel 1 south and 3 east block.

b= 4 ! / (1! * 3 !) = 4

probability = b/a = 4/20 = 1/5

Let me know if anything is not clear to you.

Hi Utam:

The question is the probaility of walking south the first 2 blocks, right? This means walking south the first block (a), and then walking south the second (b), that is:

P(a)*P(b)a = (3/6)*(2/5) = 6/30 = 1/5

Am I wrong?

Silvia

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uttam.albela: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Fri Aug 28, 2009 2:40 am; Location: Singapore; Thanked: 11 times

by uttam.albela » Thu Oct 22, 2009 8:49 pm

ssuarezo wrote:
uttam.albela wrote:Hi Abdulla,

a = To reach the destination, first find out total number of ways.
b = Then find out the number of ways in a particular manner mentioned in the problem.

Probability = b/a problem solved

Not let us find a and b.

First a:

Alicia has to cross 3 south and 3 east blocks. And, she can do this in any order.

So we need to find in how many ways we can arrange 3 south and 3 east blocks.

a = 6 ! / (3! * 3!) = 20

Now find b:

She has already traveled 2 south block, so she needs to travel 1 south and 3 east block.

b= 4 ! / (1! * 3 !) = 4

probability = b/a = 4/20 = 1/5

Let me know if anything is not clear to you.
Hi Utam:

The question is the probaility of walking south the first 2 blocks, right? This means walking south the first block (a), and then walking south the second (b), that is:

P(a)*P(b)a = (3/6)*(2/5) = 6/30 = 1/5

Am I wrong?

Silvia

Hi Silvia,

I liked your approach for getting the probability. Nice approach of getting probability of first south out of total 3 south and 3 east block and then get 2nd south out of 2 south and 3 east blocks.

My approach would have been good, if we needed to find only the number of ways of doing it. But for getting probability your approach is the best

Thanks a lot !

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sanjana: Master | Next Rank: 500 Posts; Posts: 182; Joined: Sun Aug 02, 2009 7:19 pm; Thanked: 18 times; GMAT Score:680

by sanjana » Thu Oct 22, 2009 10:15 pm

I would go the MGMAT anagram way.

We know that without any restriction she has to go 3 blocks south and 3 blocks east to get to work
Hence,we need to find the number of arrangements of the anagram SSSEEE

Total arrangements : 6!/3!*3! = 20

We know that she has to take 2 blocks south 1st. So 1st 2 are fixed SS. The remaining 4 steps we will have to find the number of arrangements of SEEE
= 4!/3!*1! = 4

Hence PRobability = No of fav/total = 4/20 = 1/5

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