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18 Jan 2013, 16:38
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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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?
(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
.
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
.
18 Jan 2013, 17:04
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HumptyDumpty
Below is given graph of y=|x|:
Attachment:
Graph.png [ 11.52 KiB | Viewed 30616 times ]
Answer: E.
PrashantPonde
Joined: 27 Jun 2012
Last visit: 29 Jan 2025
Posts: 321
Given Kudos: 185
Concentration: Strategy, Finance
Schools: Haas EWMBA '17
18 Jan 2013, 17:12
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Attachment:
XY Plane probability.jpg [ 171.15 KiB | Viewed 30256 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.
.
