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