Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 22:23

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# An (x, y) coordinate pair is to be chosen at random from the

Author Message
TAGS:

### Hide Tags

Manager
Joined: 12 Dec 2012
Posts: 160
Location: Poland
Followers: 5

Kudos [?]: 187 [1] , given: 67

An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

18 Jan 2013, 16:38
1
KUDOS
13
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

76% (02:15) correct 24% (01:10) wrong based on 406 sessions

### HideShow timer Statistics

An (x, y) coordinate pair is to be chosen at random from the xy-plane. What is the probability that $$y\geq|x|$$?

(A) 1/10
(B) 1/8
(C) 1/6
(D) 1/5
(E) 1/4

Here's how I bumped on the answer - very quickly, but I am not 100% sure if logic I used was correct. Someone could proof please?
[Reveal] Spoiler:
The |x| is always positive.

The y is positive in two quadrants - so $$P[y\geq0]=\frac{1}{2}$$.

(I am deliberately ignoring here y=0, because theoretically the axes lie inbetween the quadrants, so 0 isn't part of any quadrant equally.)

So, when $$P[y\geq0=\frac{1}{2}]$$, we need to figure out $$P[y>x]$$.

Since there is infinite number of possibilities for positive values of x and y, I concluded that there's a 50% chance of picking (x, y) at random, such that $$y>x$$.

(Here's another apparent leave-out: the possibility that $$x=y$$. However, this is only one possibility among infinity, so as for me it's acceptable. Please correct me if I am wrong.)

This way we've got $$P[y\geq0]=\frac{1}{2}$$ and $$P[y>x]=\frac{1}{2}$$.

Therefore $$P[y\geq|x|]=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$$.

The answer is correct and I hope it's not pure luck .
[Reveal] Spoiler: OA

_________________

If I answered your question with this post, use the motivating power of kudos!

Math Expert
Joined: 02 Sep 2009
Posts: 38908
Followers: 7740

Kudos [?]: 106263 [7] , given: 11618

Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

18 Jan 2013, 17:04
7
KUDOS
Expert's post
5
This post was
BOOKMARKED
HumptyDumpty wrote:
An (x, y) coordinate pair is to be chosen at random from the xy-plane. What is the probability that $$y\geq|x|$$?

(A) 1/10
(B) 1/8
(C) 1/6
(D) 1/5
(E) 1/4

Here's how I bumped on the answer - very quickly, but I am not 100% sure if logic I used was correct. Someone could proof please?
[Reveal] Spoiler:
The |x| is always positive.

The y is positive in two quadrants - so $$P[y\geq0]=\frac{1}{2}$$.

(I am deliberately ignoring here y=0, because theoretically the axes lie inbetween the quadrants, so 0 isn't part of any quadrant equally.)

So, when $$P[y\geq0=\frac{1}{2}]$$, we need to figure out $$P[y>x]$$.

Since there is infinite number of possibilities for positive values of x and y, I concluded that there's a 50% chance of picking (x, y) at random, such that $$y>x$$.

(Here's another apparent leave-out: the possibility that $$x=y$$. However, this is only one possibility among infinity, so as for me it's acceptable. Please correct me if I am wrong.)

This way we've got $$P[y\geq0]=\frac{1}{2}$$ and $$P[y>x]=\frac{1}{2}$$.

Therefore $$P[y\geq|x|]=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$$.

The answer is correct and I hope it's not pure luck .

Below is given graph of y=|x|:
Attachment:

Graph.png [ 11.52 KiB | Viewed 4884 times ]
All points which satisfy $$y\geq|x|$$ condition lie above that graph. You can see that portion of the plane which is above the graph is 1/4.

_________________
Current Student
Joined: 27 Jun 2012
Posts: 412
Concentration: Strategy, Finance
Followers: 85

Kudos [?]: 841 [4] , given: 184

Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

18 Jan 2013, 17:12
4
KUDOS
3
This post was
BOOKMARKED
Attachment:

XY Plane probability.jpg [ 171.15 KiB | Viewed 4886 times ]

Just plot the y>|x| on your paper and you will see it takes 1/4 of the space i.e. Probability = 1/4

_________________

Thanks,
Prashant Ponde

Tough 700+ Level RCs: Passage1 | Passage2 | Passage3 | Passage4 | Passage5 | Passage6 | Passage7
VOTE GMAT Practice Tests: Vote Here
PowerScore CR Bible - Official Guide 13 Questions Set Mapped: Click here

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15474
Followers: 649

Kudos [?]: 209 [0], given: 0

Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

09 Jun 2014, 08:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Senior Manager
Joined: 17 Sep 2013
Posts: 388
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Followers: 19

Kudos [?]: 296 [1] , given: 139

An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

20 Dec 2014, 06:45
1
KUDOS
What I did -->

We know that we have to choose a point such that Y is positive--> so while choosing any number from the number line..the probability of selecting a number >0 or <0 should be equal..So Prob of choosing a positive Y is 1/2.
Say a coordinate x is chosen...Now,The probability of choosing any number that is <x or >x is again equal..so the Probability is again 1/2 for each case.

Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4.

Can any expert vouch for this way of solving such a question??
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..

Director
Joined: 07 Aug 2011
Posts: 579
GMAT 1: 630 Q49 V27
Followers: 3

Kudos [?]: 449 [0], given: 75

An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

04 Jan 2015, 03:10
JusTLucK04 wrote:
What I did -->

We know that we have to choose a point such that Y is positive--> so while choosing any number from the number line..the probability of selecting a number >0 or <0 should be equal..So Prob of choosing a positive Y is 1/2.
Say a coordinate x is chosen...Now,The probability of choosing any number that is <x or >x is again equal..so the Probability is again 1/2 for each case.

Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4.

Can any expert vouch for this way of solving such a question??

Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4.
this is simply wrong as you are reading P(y>=|x|) as P(y>|x| AND y=|x|) this should be 0.
question is asking P(y>=|x|) which is P(y>|x| ) OR P(y=|x|) -> P(y>|x| ) + P(y=|x|)
_________________

Thanks,
Lucky

_______________________________________________________
Kindly press the to appreciate my post !!

Senior Manager
Joined: 15 Sep 2011
Posts: 362
Location: United States
WE: Corporate Finance (Manufacturing)
Followers: 7

Kudos [?]: 338 [0], given: 45

An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

30 Jun 2015, 19:12
JusTLucK04 wrote:
What I did -->

We know that we have to choose a point such that Y is positive--> so while choosing any number from the number line..the probability of selecting a number >0 or <0 should be equal..So Prob of choosing a positive Y is 1/2.
Say a coordinate x is chosen...Now,The probability of choosing any number that is <x or >x is again equal..so the Probability is again 1/2 for each case.

Now for P(y) such that both the conditions are met will be 1/2*1/2= 1/4.

Can any expert vouch for this way of solving such a question??

Had there been constraints, such as $$y\geq{|\frac{1}{2}x|}$$, or say $$y\geq{x^{2}}$$, the range would vary enough to where $$\frac{1}{2} * \frac{1}{2}$$ estimation would be inaccurate, and therefore the solution above would not have worked. You might have adjusted for it, but since the question is out there, I took a swipe. Cheers
Intern
Joined: 07 Apr 2016
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 15

Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

14 Apr 2016, 20:30
this is how i did it -

without any restrictions, you had 2 choices of y (+/-) and 2 for x. a total of 4 choices. now you just have 1 for y. if you take quadrant I, half the values of y will be > x and half of the values of x>y's. so prob of y=>x = 1/2. but x can be -ve as well as we're comparing absolute values. so the prob of y=>lxl is 1/2*1/2 = 1/4
Intern
Joined: 10 Oct 2016
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 9

Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

09 Jan 2017, 11:04
PrashantPonde wrote:
Attachment:
XY Plane probability.jpg

Just plot the y>|x| on your paper and you will see it takes 1/4 of the space i.e. Probability = 1/4

Another way of reaching the same conclusion is as follows:

The area within the function y=lxl subtends an angle 90 deg with the coordinate system. Since the lines y=x and y=-x have slopes +45 and -45 respectively.

The total angle of any coordinate system is 360 deg. Therefore, the probability that the point lies within the 90 deg region is 90/360 = 1/4
Senior Manager
Joined: 25 Mar 2013
Posts: 276
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
Followers: 3

Kudos [?]: 29 [0], given: 101

Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]

### Show Tags

19 Jan 2017, 17:00
X,Y random
Plug In values with different signs both -3 and -6
1, 4 > 3
2, 2 > 3
3, -5 > 3
4, -5 > 6
Only case 1 is right
1/4 probability
Is this a right approach however i solved in 2:40sec
E
_________________

I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

Re: An (x, y) coordinate pair is to be chosen at random from the   [#permalink] 19 Jan 2017, 17:00
Similar topics Replies Last post
Similar
Topics:
1 Two numbers, X and Y, are to be chosen at random from the set of numbe 1 07 Apr 2016, 17:11
40 If x is to be chosen at random from the set {1, 2, 3, 4} and 27 22 Oct 2016, 11:25
5 If x and y are both integers chosen at random between 1 and 3 19 Mar 2017, 03:19
If x is to be chosen at random from the set {1,2,3,4} and y 3 17 Feb 2014, 02:21
29 If x is to be chosen at random from the set {1, 2, 3, 4} and 14 27 Mar 2017, 11:49
Display posts from previous: Sort by