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An (x, y) coordinate pair is to be chosen at random from the
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18 Jan 2013, 15:38
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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 .
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Re: An (x, y) coordinate pair is to be chosen at random from the
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18 Jan 2013, 16:04
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 9686 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.
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Re: An (x, y) coordinate pair is to be chosen at random from the
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18 Jan 2013, 16:12
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 Hence choice(E) is the answer.
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An (x, y) coordinate pair is to be chosen at random from the
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20 Dec 2014, 05:45
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??



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An (x, y) coordinate pair is to be chosen at random from the
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04 Jan 2015, 02:10
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)



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An (x, y) coordinate pair is to be chosen at random from the
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30 Jun 2015, 18:12
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



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Re: An (x, y) coordinate pair is to be chosen at random from the
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14 Apr 2016, 19:30
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



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Re: An (x, y) coordinate pair is to be chosen at random from the
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09 Jan 2017, 10:04
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



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Re: An (x, y) coordinate pair is to be chosen at random from the
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19 Jan 2017, 16:00
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



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Re: An (x, y) coordinate pair is to be chosen at random from the
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19 Aug 2019, 20:37
Bunuel wrote: 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 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. 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?



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Re: An (x, y) coordinate pair is to be chosen at random from the
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19 Aug 2019, 22:00
subhasisray85 wrote: Bunuel wrote: 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: The attachment Graph.png is no longer available 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. 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°.
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Re: An (x, y) coordinate pair is to be chosen at random from the
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06 Jan 2020, 13:26
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
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06 Jan 2020, 13:26






