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?


1 Answers

David Shabazi Profile
David Shabazi answered

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.
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John McCann
John McCann commented
" integers "

Without a fraction or decimal component..

So, 13/81 would be rather strange here given the problem..
Rojer Baroi
Rojer Baroi commented
OK let me say the answer in the graphs the possibilities would be from -4 to 4 for 'x' and 'y'. After that there would be 81 possibilities.Out of them 13 of them would be at least 2.This really fascinating problem. Try it yourself.I was fortunately able to do it.
John McCann
John McCann commented
13/81 is not how you would express...." After that there would be 81 possibilities.Out of them 13 of them would be at least 2.T ".....that! Even if you were correct.

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