jainrahul1985 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) 1/4 b) 1/3 c) 1/2 d) 3/5 e) 2/3

OA A

Can someone please explain how to approach this question in the easiest way .

x^2+y^2=1; represents equation of circle whose center lies at the origin (0,0) and having radius =1;

if we draw the circle considering origin as the center we will find out that it touches the XY axis at the point (1,0),(-1,0),(0,1),(0,-1) of these values;

Now looking at the conditions b>a+1; only portion of area where this thing can be achieved lies between the points (-1,0) and (0,1), for all other portions points will lie outside the circle,

Therefore of all the 4 portions (i.e. we are dividing circle into 4 parts (I,II,III,IV) quadrant) only second quadrant points yields the desired result hence porbability is 1/4..!!

I hope it helps..!!!

jainrahul1985 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) 1/4 b) 1/3 c) 1/2 d) 3/5 e) 2/3

OA A

Can someone please explain how to approach question in the easiest way .

The easiest approach is to draw the circle and the line y=x+1:



Every point (a,b) above the line y=x+1 satisfies the given condition that b > a+1. As shown above, a quarter of the circle (the portion shaded in blue) will lie above y=x+1. Thus, the probability is of picking a point (a,b) on the circle such that b>a+1 is 1/4.

The correct answer is A.

awesome approach by both experts . you made this question so simple for me . thanks a ton

Are a,b integers here??