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An (x, y) coordinate pair is to be chosen at random from the [#permalink]
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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 [#permalink]
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18 Jan 2013, 16:04
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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 5798 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 [#permalink]
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18 Jan 2013, 16:12
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Attachment:
XY Plane probability.jpg [ 171.15 KiB  Viewed 5799 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 [#permalink]
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20 Dec 2014, 05:45
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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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 [#permalink]
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