Problem Solving

If point (x, y) is selected at random from a square region that has vertex (0, 0), (2, 0), (0, 2), and (2, 2), what is the probability that y-x>1?
(A) 1/8
(B) 3/8
(C) 1/4
(D) 5/8
(E) 3/4

Is it C?

Probability of y > x + 1

0 > 0 + 1 not true

0 > 2 + 1 not true

2 > 0 + 1 true

2 > 2 + 1 not true

1 out of 4 or 1/4.

Cant x & y have any value in between 0-2 if randomly selected?

i am confused !

You bring up a good point ksh, I might have gone for the easy (trap) answer choice.

Hope someone else could shed some light on this one.

Hi all,

Well, y=x+1 is a line. So y>x+1 defines a region above that line.

If you draw the square region and the line you gonna see that the line intersects the Y axis in y=1 and the square top line in the point (1,2) forming a triangle of side 1.

So the probability is the area of the triangle divided by the area of the square.

Area of the triangle = 1/2

Area of the square = 4

So probability is (1/2)/4 = 1/8, chose A.

Answer is 1/8

Points are

0,0
0,1
0,2
1,0
2,0
2,1
2,2
1,1

1/8.

is 1/8 the answer

if that is the answer the solution is

y-x =1 is the line which passes through points (1,2) and (0,1) and y-x>1 corresponds to area above this line

the total area of the square is 4

the area of the region above the line y-x=1 is 1/2*1*1 = 1/2

hence the probability is 1/2 divided by the whole area which is 4

hence 1/8

atlantic..its right...amazing explanation..