Is the answer 1/3. Could you post answer choices?

The answer is [spoiler]1/3[/spoiler]
Look at the attached figure. We can infer that this right triangle is in the 1st quadrant. Anywhere in the first quadrant, below the line y=x the x co-ordinates will always be greater than y. The line y=x also divides the triangle in two parts. So the probability we are looking for is
Area of the lower part of the triangle/Total area of the triangle

Area of the whole triangle=(1/2)*10*5

Now, if a line has end points 0,10 and 5,0 its equation can be found out using the formula
(y-y1)/(x-x1)=(y1-y2)/(x1-x2)

Substituting we get 2x+y=10. Now, this line intersects with y=x so if we solve both simultaneously we get x=10/3 and y=10/3 which are the co-ordinates of the point of intersection of both lines.

Now from the point(10/3,10/3) drop a perpendicular to x-axis and this perpendicular would have co-ordinates (10/3,0) . Because of this perpendicular, the lower triangle is divided into two smaller triangles. We need to find the sum of the areas of these two smaller triangles.

For the triangle on the left
Area=(1/2)*(10/3)*(10/3)

For triangle on the right
Area=(1/2)*(5-10/3)*(10/3)

Adding both the above we get
Combined area=50/6=Area of the lower part of the triangle

Area of the lower part of the triangle/Total area of the triangle=
(50/6)/[(1/2)*5*10]=2/6=[spoiler]1/3 [/spoiler]

What is the OA?

If my understanding is correct, if you divide the right triangle in to two triangles of equal area. Only in one half of the right triangle will you have points where y<x or y>x

Area of this right triangle is 25. One half of this area is 12.5

Therefore, the probability that y<x is 12.5/25 = 1/2, this probability also includes those points where y=x

Me did the exact thing, guess its correct.

knight247 wrote:The answer is [spoiler]1/3[/spoiler]

@sl750
Yes, IF the line y=x was dividing the triangle into two halves then the probability that we are looking for would be 1/2. But clearly in this case the line y=x is not dividing the triangle into two halves. 1/2 is not the answer for this problem.

sushbis wrote:A triangle has following coordinates (0,0),(0,10) and (5,0). What is the probability that point x,y in this right angled triangle has y<x?

Draw the said triangle OAB with O (0, 0), A (0, 10) and B (5, 0) on the coordinate axis first. Obtain the equation of line AB as x + 2 y = 10. Then draw the line y = x that connects O to the line AB in C (10/3, 10/3). Now all points in the triangle OCA are such that y > x and all points in the triangle OCB are such that y < x.

Total area of triangle OAB = ½ × 10 × 5 = 25, and area of triangle OCB = ½ × 5 × 10/3 = 25/3

and hence the required probability = (25/3)/25 = [spoiler]1/3[/spoiler]