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In the cartesian plane, selection of a point P along the y axis in [0,2] is uniformly random. Similarly selection of a point Q along the x axis in [0,2] also uniformly distributed. What is the probability of the area of the triangle POQ to be less than or equal to 1, where O is the origin ?
I am getting 0.5

Area of POQ=$\frac{1}{2}*2*2=2$

Area less than or equal to 1 is $\frac{1}{2}*1*2=1$

So, Probability of POQ less than equal to 1 is $\frac{1}{2}=0.5$

You have not considered the part x = [1,2] where y can be in range [0,$\frac{2}{x}$]

Solution:

we need $\frac{xy}{2} \leq 1$ Or, $xy \leq 2$

Just take the area under this rectangular hyperbola bounded by four lines:

y=0,x=0,x=2,y=2

Required area = $\frac{1+\ln (2)}{2}$

=>P = 0.846

Alternatively,

We can do by the integration method in two parts

• first, x from zero to 1 => p = 0.5 [ valid y range here is [0,2] ]
• second , x from 1 to 2 => p = 0.346 [ valid y range here is [0,2/x] ]
• P = 0.846
Can you Please Explain the the Integration part i.e. how you are getting the values?