Probablity

[vipulgoyal]

2. Point (x,y) is a point within the triangle. What is the probability that y<x?

PFA the digram

Ans not given in source

My take on it 1/2

[Anju@Gurome]

Refer to the figure below,


For y < x, the point (x, y) must lie below the line y = x on the xy coordinate plane.
Hence, here the point (x, y) must be inside the triangle OMB.

Hence, required probability = (Area of OMB)/(Area of OAB)

Area of OAB = (base)*(height)/2 = 5*10/2
Area of OMB = (base)*(height)/2 = 5*(y-coordinate of M)/2

Hence, required probability = (y-coordinate of M)/10

Now, M is the intersection of line segment AB and x = y.
Equation of the line segment AB : x/5 + y/10 = 1
If y-coordinate of M is Y, then Y/5 + Y/10 = 1 ---> 3Y/10 = 1 ---> Y = 10/3

Hence, required probability = (10/3)/10 = 1/3

[Anju@Gurome]

Another approach to solve this problem without finding the coordinates of M is as follows.

Note that M lies on the x = y.
Hence, x-coordinate of M = y-coordinate of M = p (say)
Required probability = (Area of OMB)/(Area of OAB) = (Area of OMB)/(Area of OMA + Area of OMB)

Now, area of OMA = (base)*(height)/2 = OA*(x-coordinate of M)/2 = 10p/2
And, area of OMB = (base)*(height)/2 = OB*(y-coordinate of M)/2 = 5p/2

Hence, required probability = (5p/2)/(10p/2 + 5p/2) = 5p/(10p + 5p) = 5/15 = 1/3