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Standing on the origin of an xy-coordinate plane, John takes

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Standing on the origin of an xy-coordinate plane, John takes  [#permalink]

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New post 13 Feb 2012, 21:36
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Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again?

(A) 7/64
(B) 9/64
(C) 11/64
(D) 13/64
(E) 15/64
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Re: Probability PS  [#permalink]

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New post 14 Feb 2012, 00:25
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Smita04 wrote:
Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again?

(A) 7/64
(B) 9/64
(C) 11/64
(D) 13/64
(E) 15/64


Find the total number of possibilities first. He takes total 4 steps . He can take each step in any direction so there are a total of 4*4*4*4 possibilities (this includes UUUU, UDLR, DDLR etc etc)

He needs to be at the origin after 4 steps. So if he takes a step up, he needs to take a step down at some time. If he takes a step to the left, he needs to take one to the right at some time. Say if he takes two steps in this way - UL, his next two steps are defined - DR/RD. If instead, he takes two steps in this way - UU, his next two steps have to be DD.
There are two possibilities:
1. He goes only Up and Down or only Left and Right. UUDD can be arranged in 4!/(2!*2!) ways (includes UDUD, DUDU, DDUU etc). LLRR can also be arranged in 4!/(2!*2!) ways.
2. He goes Up/Down as well as Left/Right. UDLR can be arranged in 4! ways.

Total possible arrangements = 2*4!/(2!*2!) + 4!

Probability he comes back to the origin = (2*4!/(2!*2!) + 4!)/4*4*4*4 = 9/64
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Re: Probability PS  [#permalink]

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New post 13 Feb 2012, 22:18
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This is the Manhattan Gmat Problem of the week this week. Extremely difficult to explain this without drawing. But lets suppose he takes a step in any one direction. Since he can take 4 different directions from there on in and he does this 3 times the total number of possibilities is \(4*4*4=64\)

Now if you start drawing on a piece of paper, you will realise that there are \(9\) such possibilities where he can end up back on the origin so answer should be \(\frac{9}{64}\). I am going to attach an image of all these possibilities along-with this post as well.

Attachment:
routes.jpg
routes.jpg [ 90.8 KiB | Viewed 3463 times ]


Hence B
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Re: Standing on the origin of an xy-coordinate plane, John takes  [#permalink]

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New post 20 May 2014, 23:57
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It took me about 3.5 mins to solve...I forgot some cases initially

Total no of cases=4^4
Way 1- 1 each of L,R,U,D- They can be arranged in 4! ways..4!
Way 2- 2 each of R & L..RRLL- They can be arranged in 4!/2!*2!= 6
Way 2- 2 each of D & U..DDUU- They can be arranged in 4!/2!*2!= 6

36 ways possible/4*4*4*4
=9/64

I think it helps to think directions as numbers with opposite signs...In this case the sum of the 4 numbers should be 0
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Re: Standing on the origin of an xy-coordinate plane, John takes  [#permalink]

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Re: Standing on the origin of an xy-coordinate plane, John takes &nbs [#permalink] 30 Nov 2018, 00:10
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