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Q.
2 positive real numbers $x \&$ y satisfy $x \leq 1 \& y \leq 1$ are chosen at random. The probability that $x+y \leq 1$, given that $x^2+y^2 \geq 1 / 4$ is
Probability - Part 2
Solution:
$n ( S )=1 ; A : x + y \leq 1 ; B : x ^2+ y ^2 \geq \frac{1}{4}$
$P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{2}-\frac{\pi}{4} \frac{1}{4}}{1-\frac{\pi}{16}}=\frac{8-\pi}{16-\pi} $
