dczuchta wrote:

Would someone please explain this problem to me? Thank you.

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 \gt a + 1 ?

(A)

\frac{1}{4}(B)

\frac{1}{3}(C)

\frac{1}{2}(D)

\frac{3}{5}(E)

\frac{2}{3}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.

I did not understand this part "You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4." Above the line and inside the boundary of the circle is just small part. How is it 1/4 of the circle