GATE | GATE-CS-2004 | Question 80

A point is randomly selected with uniform probability in the X-Y plane within the rectangle with corners at (0,0), (1,0), (1,2) and (0,2). If p is the length of the position vector of the point, the expected value of p2 is

(A) 2/3
(B) 1
(C) 4/3
(D) 5/3


Answer: (D)

Explanation: Here minimum value of p can be 0 (if point chosen is (0,0), then length of position vector will be 0), and maximum value can be 5√ when point chosen is (1,2), because that is the farthest point from origin. So p can vary from 0 to 5√.
Now we know that

E(p^2) = \int^\sqrt{5}_0 p^2*P(p)\,dp

Since p is a uniform random variable, P(p) = \frac{1}{\sqrt{5}-0} = \frac{1}{\sqrt{5}}
So
E(p^2) = \frac{1}{\sqrt{5}}\left[\frac{p^3}{3}\right]^{\sqrt{5}}_0 = \frac{5}{3}
So option (D) is correct.

Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2004.html

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