Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 12 Dec 2012
Posts: 154
Location: Poland

An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
Show Tags
18 Jan 2013, 16:38
Question Stats:
77% (01:13) correct 23% (01:03) wrong based on 461 sessions
HideShow timer Statistics
An (x, y) coordinate pair is to be chosen at random from the xyplane. What is the probability that \(y\geqx\)? (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? 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 leaveout: 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 \(P[y\geqx]=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}\). The answer is E. The answer is correct and I hope it's not pure luck .
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
If I answered your question with this post, use the motivating power of kudos!




Math Expert
Joined: 02 Sep 2009
Posts: 47983

Re: An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
Show Tags
18 Jan 2013, 17:04
HumptyDumpty wrote: An (x, y) coordinate pair is to be chosen at random from the xyplane. What is the probability that \(y\geqx\)? (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? 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 leaveout: 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 \(P[y\geqx]=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}\). The answer is E. 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 6500 times ]
All points which satisfy \(y\geqx\) condition lie above that graph. You can see that portion of the plane which is above the graph is 1/4. Answer: E.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Senior Manager
Joined: 27 Jun 2012
Posts: 390
Concentration: Strategy, Finance

Re: An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
Show Tags
18 Jan 2013, 17:12
Attachment:
XY Plane probability.jpg [ 171.15 KiB  Viewed 6502 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 Hence choice(E) is the answer.
_________________
Thanks, Prashant Ponde
Tough 700+ Level RCs: Passage1  Passage2  Passage3  Passage4  Passage5  Passage6  Passage7 Reading Comprehension notes: Click here VOTE GMAT Practice Tests: Vote Here PowerScore CR Bible  Official Guide 13 Questions Set Mapped: Click here Looking to finance your tuition: Click here




Retired Moderator
Joined: 17 Sep 2013
Posts: 369
Concentration: Strategy, General Management
WE: Analyst (Consulting)

An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
Show Tags
20 Dec 2014, 06:45
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: 560
Concentration: International Business, Technology

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: 342
Location: United States
WE: Corporate Finance (Manufacturing)

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: 1

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

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 Hence choice(E) is the answer.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: 263
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5

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



NonHuman User
Joined: 09 Sep 2013
Posts: 7759

Re: An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
Show Tags
25 Jan 2018, 13:01
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: An (x, y) coordinate pair is to be chosen at random from the &nbs
[#permalink]
25 Jan 2018, 13:01






