Hi, I didn't know how to post a new answer for https://web2.0calc.com/questions/a-point-x-y-is-randomly-selected-such-that-0-le-x-le-8-and-0-le-y-le-4-what-is-the-probability-that-x-y-le-4-express-your-a , but it seems like the answer you put for this was wrong. You said that it was 1/4 (8/32). However, I think that there are actually 9 choices for x and 5 choices for y, meaning 45 choices total. Out of these 45, 15 meet the requirement. Therefore, the answer is 1/3. Is that right? Thank you.

Guest Jan 12, 2019

#1**+2 **

Mmmmm.....the area of the feasible region is 8 * 4 = 32 units^2

x + y ≤ 4 will form a triangle in this region with a base and height of 4.........so its area is 8

So....the probability of a point falling into this area of the feasible region = 8 / 32 = 1 / 4

CPhill Jan 12, 2019

#2**+2 **

This problem requires integer coordinates, so it is not a geometric probability problem.

There are 9 choices for x and 5 choices for y so there are 9*5=45 possible choices for (x,y). The pairs for which x+y ≤ 4 are

(0,0) (0,1) (1,0) (0,2) (1,1) (2,0) (0,3) (1,2) (2,1) (3,0) (0,4) (1,3) (2,2) (3,1) (4,0)

These are 15 of the 45 possible choices, so the probability is 15/45, or 1/3. I still don't understand your method, could you explain it to me, please?

Guest Jan 12, 2019

edited by
Guest
Jan 12, 2019