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

Question

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?

Show Spoiler

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

.

Bunuel · Accepted Answer

Below is given graph of y=|x|: 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.

PrashantPonde · Answer

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.

JusTLucK04 · Answer

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??

Lucky2783 · Answer

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|)

mejia401 · Answer

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

achalbhatt · Answer

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

seneca04 · 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

kanusha · Answer

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

GMAC700 · Answer

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?

Bunuel · Answer

The whole plane is 360° and the portion of the plane which is above the graph is 90°, which is 1/4 of 360°.

janadipesh · Answer

<you had 2 choices of y (+/-) and 2 for x> ....what about the point 0 ?

waytowharton · Answer

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?

Bunuel · Answer

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.

Elite097 · Answer

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

avigutman · Answer

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.

bumpbot · Answer

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