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