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13 Feb 2012, 22:36
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (02:46) correct
59%
(02:42)
wrong
based on 321
sessions
History
Date
Time
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Not Attempted Yet
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
(A) 7/64
(B) 9/64
(C) 11/64
(D) 13/64
(E) 15/64
14 Feb 2012, 01:25
Kudos
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Smita04
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
13 Feb 2012, 23:18
Kudos
Bookmarks
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.
routes.jpg [ 90.8 KiB | Viewed 9616 times ]
Hence B
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 [ 90.8 KiB | Viewed 9616 times ]
Hence B
