We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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