2 positive real numbers x y satisfy x ≤ 1 y ≤ 1 are chosen at random. The probability that x+y ≤ 1, given that x2+y2 ≥ 1 / 4 is

Question

2 positive real numbers x y satisfy x ≤ 1 y ≤ 1 are chosen at random. The probability that x+y ≤ 1, given that x2+y2 ≥ 1 / 4 is (A) (8/16-π) (B) (4-π/16-π) (C) (8-π/16-π) (D) none

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b0a84de37100e325f1124b7eaed9fc8d-.png" /> n ( S )=1 ; A: x + y ≤ 1 ; B: x 2+ y 2 ≥ (1/4) P(A / B)=(P(A ∩ B)/P(B))=((1/2)-(π/4) (1/4)/1-(π)16)=(8-π/16-π)

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

$\frac{8}{16 - π}$

$\frac{4 - π}{16 - π}$

$\frac{8 - π}{16 - π}$

none

$n (S) = 1; A : x + y \leq 1; B : x^{2} + y^{2} \geq \frac{1}{4}$
$P (A / B) = \frac{P ( A \cap B )}{P ( B )} = \frac{\frac{1}{2} - \frac{π}{4} \frac{1}{4}}{1 - \frac{π}{16}} = \frac{8 - π}{16 - π}$

Q. 2 positive real numbers x& y satisfy x≤1&y≤1 are chosen at random. The probability that x+y≤1, given that x2+y2≥1/4 is

16−π8​

16−π4−π​

16−π8−π​