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

Question

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

KarishmaB · Accepted Answer

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

omerrauf · Answer

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 [ 90.8 KiB | Viewed 9608 times ]

Hence B

JusTLucK04 · Answer

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

MissionAdmit · Answer

Total ways to fill 4 steps with 4 directions ( Order matters) = 4P1 x 4P1 x 4P1 x 4P1= 4x4x4x4

Now Favourable Ways in which we can reach the origin after 4 steps starting from the origin= Scenario 1 + Scenario 2

Scenario 1) We take a different direction in each step - no direction is repeated = 4P1 x 3P1 x 2P1 x 1P1 =4x3x2x1

Scenario 2) We take only two directions in 4 steps such as we return to the origin, which means we can use any one direction only twice. Two sets of directions can fill up these four steps 
a) Only R and L ( 2 steps max with R and 2 Steps max with L) = 4P2 MINUS when one direction repeats in more than 2

RRLL
RLRL
RLLR
LLRR
LRLR
LRRL

6 steps

b) Only T and D ( 2 steps max with T and 2 Steps max with D) = 6 Steps following the above approach

Thus a+b = 12

Scenario 1 + Scenario 2 = 24+12= 36

P ( Favourable/ Total) = 36/256= 9/64

Kinshook · Answer

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?

For every step, there are 4 possibilities.
Therefore, total ways = 4^4 = 256

For his to return back to origin, 2 steps will have corresponding opposite direction steps.
Favorable ways = 4^2 = 16

Directions = {U, D, L, R}
Number of ways of type UDLR = 4! = 24
Number of ways of type UUDD or LLRR = 2* 4!/2!2! = 12
Total favourable ways = 24 + 12 = 36

The probability that he winds up at the origin again = 9/64

IMO B

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

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GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards