# There is a point in the plane with coordinates (x,y),where x and y are integers having numerical value less than or equal to four. What is the probability that the distance of the point from the origin is at most two units? Really need help! The answer is suppose to be 13/81. Can someone please show the way though?

The distance formula will be involved in this question, which is:

D = sqrt{(x2 - x1)^2 + (y2 - y1)^2}

Two points will be used to plug into this equation: The origin, which is (0, 0) and (x, y).

Since the first point is the origin, then both zeros are x1 and y1, so the distance formula can be rewritten for this problem as:

D = sqrt{(x2)^2 + (y2)^2}

x2 and y2 represent (x, y) respectively. So now you just need to figure out numbers for both x and y to plug into the distance formula that will give a distance of 2 units or less. For instance, if we were to plug in (sqrt(2), sqrt(2)) for (x, y), that would give us 2.

D = sqrt{ [sqrt(2)]^2 + [sqrt(2)]^2 } = sqrt(2 + 2) = sqrt(4) = 2

Is the answer really 13/81? That seems a little bit strange for this problem, but I'll get back to it after I finish school.

thanked the writer.