Region R is defined as region in first quadrant satisfying the condition x^2 ...

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Question Id : 11663 | Context :NIMCET 2024

Region R is defined as region in first quadrant satisfying the condition

$x^2 + y^2 < 4$ . Given that a point P=(r,s) lies in R, what is the probability that r>s?

🎥 Video solution / Text Solution of this question is given below:

Given: $R = \{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 4 \}$ in the first quadrant

Area of region $R$ in first quadrant:

$A = \frac{1}{4} \pi (2)^2 = \pi$

Region where $r > s$ (i.e., below line $x = y$ ) occupies half of that quarter-circle:

$A_{\text{favorable}} = \frac{1}{2} \pi$

Therefore, the required probability is:

$\text{Probability} = \frac{\frac{1}{2} \pi}{\pi} = \boxed{\frac{1}{2}}$

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