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An (x, y) coordinate pair is to be chosen at random from the

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An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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New post 18 Jan 2013, 15:38
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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 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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New post 18 Jan 2013, 16:04
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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 E.

The answer is correct and I hope it's not pure luck :).


Below is given graph of y=|x|:
Attachment:
Graph.png
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.

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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New post 18 Jan 2013, 16:12
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Attachment:
XY Plane probability.jpg
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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New post 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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New post 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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New post 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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New post 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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New post 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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New post 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
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Re: An (x, y) coordinate pair is to be chosen at random from the  [#permalink]

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New post 19 Aug 2019, 20:37
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 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 \(y\geq|x|\) 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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New post 19 Aug 2019, 22:00
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 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 \(y\geq|x|\) 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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Untitled.png [ 13.3 KiB | Viewed 1341 times ]


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

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New post 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 ?
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Re: An (x, y) coordinate pair is to be chosen at random from the   [#permalink] 06 Jan 2020, 13:26
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