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An (x, y) coordinate pair is to be chosen at random from the [#permalink]
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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]
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]
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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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]
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]
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]
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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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]
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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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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]
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]
Bunuel - Would the answer change if we did not have the equal to sign i.e. y>|x|? If no, then why because some points would get added?
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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]
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waytowharton wrote:
Bunuel - Would the answer change if we did not have the equal to sign i.e. y>|x|? If no, then why because some points would get added?


The answer will still be exactly 1/4, no matter whether the question says \(y\geq|x|\) or \(y>|x|\). That's because a line is one-dimensional: a line segment can only have length but no area (the same way a point has no dimensions, only position). So, since the region where \(y=|x|\) is just line segments (check red graph in my solution), then its area is zero.
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Re: An (x, y) coordinate pair is to be chosen at random from the [#permalink]
avigutman how to solve traditionally if we do not know graphs as i generally have not learnt specifically tricks to plot graphs especially for mods and number substitution is not handy at most times
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An (x, y) coordinate pair is to be chosen at random from the [#permalink]
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Elite097 wrote:
avigutman how to solve traditionally if we do not know graphs as i generally have not learnt specifically tricks to plot graphs especially for mods and number substitution is not handy at most times

I don't think there's any need for graphs here, Elite097, or even pen and paper in general.
Here's how I reason through this question: as a first step, I ask myself what's the probability that y is greater than x. Since they're both chosen at random, that's going to be exactly half (there's an equal chance for each of them to be greater than the other).
with no integer constraint, "greater than" is the same thing as "greater or equal"
Next, I address the fact that x is in absolute value in the question stem. What impact does that have? It cuts in half the possible values of x. What does this mean for the previous step in my reasoning? All those cases in which y was greater than x... half of those cases no longer exist (those cases in which x had negative values).
Therefore, the answer is half of half, or 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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