Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 27 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 12 Dec 2012
Posts: 151
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% (01:40) correct 23% (01:34) wrong based on 389 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: 56306

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 8151 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.
_________________




Senior Manager
Joined: 27 Jun 2012
Posts: 364
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 8150 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  Passage7Reading Comprehension notes: Click hereVOTE GMAT Practice Tests: Vote HerePowerScore CR Bible  Official Guide 13 Questions Set Mapped: Click hereFinance your Student loan through SoFi and get $100 referral bonus : Click here




Senior Manager
Joined: 17 Sep 2013
Posts: 329
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: 518
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)



Senior Manager
Joined: 15 Sep 2011
Posts: 319
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



Manager
Joined: 25 Mar 2013
Posts: 235
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: 11719

Re: An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
Show Tags
23 Mar 2019, 10:58
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.
_________________




Re: An (x, y) coordinate pair is to be chosen at random from the
[#permalink]
23 Mar 2019, 10:58






