The X- and y-coordinates of point P are each to be chosen at random from the set of integers 1 through 10. What is the probability that P will be in quadrant II ?
О
1/10
1/4
1/2

Ok, as i understand it:

for a point P = (x, y)

The values of x and y can be randomly chosen from the set {1, 2, ..., 10}

We want to find the probability that the point P lies on the second quadrant:

First, what type of points are located in the second quadrant?

We should have a value negative for x, and positive for y.

But in our set;  {1, 2, ..., 10}, we have only positive values.

So x can not be negative, this means that the point can never be on the second quadrant.

So the probability is 0.

ruth is 6 years old

step-by-step explanation:

6 + 24 = 30,   so 6/30 = .2  

if you do 1/5 it also equals .2, so that is the answer.