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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # An (x, y) coordinate pair is to be chosen at random from the

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Manager  Joined: 12 Dec 2012
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Location: Poland
An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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Difficulty:   15% (low)

Question Stats: 77% (01:36) correct 23% (01:33) wrong based on 491 sessions

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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?
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 .
Math Expert V
Joined: 02 Sep 2009
Posts: 61396
Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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10
6
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?
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 9686 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.

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Senior Manager  Joined: 27 Jun 2012
Posts: 347
Concentration: Strategy, Finance
Schools: Haas EWMBA '17
Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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7
1
2
Attachment: XY Plane probability.jpg [ 171.15 KiB | Viewed 9686 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

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Prashant Ponde

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##### General Discussion
Retired Moderator B
Joined: 17 Sep 2013
Posts: 316
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An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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1
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??
Senior Manager  Joined: 07 Aug 2011
Posts: 492
GMAT 1: 630 Q49 V27
An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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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: 302
Location: United States
WE: Corporate Finance (Manufacturing)
An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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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]

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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
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Joined: 10 Oct 2016
Posts: 2
Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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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
Manager  S
Joined: 25 Mar 2013
Posts: 221
Location: United States
Concentration: Entrepreneurship, Marketing
GPA: 3.5
Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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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
Intern  B
Joined: 26 Aug 2014
Posts: 1
Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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Bunuel wrote:
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?
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
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.

Hello Bunuel,

I couldn't able to understand how to reached to 1/4 from the graph. Can you please elaborate more from the graph on how you derived at 1/4 probability?
Math Expert V
Joined: 02 Sep 2009
Posts: 61396
Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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2
subhasisray85 wrote:
Bunuel wrote:
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?
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:
The attachment Graph.png is no longer available
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.

Hello Bunuel,

I couldn't able to understand how to reached to 1/4 from the graph. Can you please elaborate more from the graph on how you derived at 1/4 probability?

The whole plane is 360° and the portion of the plane which is above the graph is 90°, which is 1/4 of 360°.
Attachments Untitled.png [ 13.3 KiB | Viewed 1341 times ]

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Concentration: Operations, Finance
WE: Manufacturing and Production (Energy and Utilities)
Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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achalbhatt wrote:
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

<you had 2 choices of y (+/-) and 2 for x>....what about the point 0 ? Re: An (x, y) coordinate pair is to be chosen at random from the   [#permalink] 06 Jan 2020, 13:26
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