mirhaque wrote:
Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?
(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}
PLS DRAW & ILLUSTRATE:
Look at the diagram below.
Attachment:
graph.php.png
The circle represented by the equation
x^2+y^2 = 1 is centered at the origin and has the radius of
r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book:
math-coordinate-geometry-87652.html ).
So, set T is the circle itself (red curve).
Question is: if point (a,b) is selected
from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.
Answer: A.
If it were: set T consists of all points (x,y) such that
x^2+y^2<1 (so set T consists of all points
inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?
Then as the area of the segment of the circle which is above the line is
\frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4} so
P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}.
Hope it's clear.
Can you pls explain how did u calculate area of segment above the line (pi-2/4) for x^2+y^2<1