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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.

 Jan 12, 2019
 #1
avatar+98172 
+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

 

 

 

 

cool  cool cool

 Jan 12, 2019
 #2
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+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?

 Jan 12, 2019
edited by Guest  Jan 12, 2019
 #3
avatar+98172 
+1

Your problem specifies "point"......so my answer is correct based on that....

However.....it should specify "lattice points" [ points with integer coordinates ]

 

Taking this into cosideration.....then 1/3 is correct......smiley smiley smiley

 

 

cool cool cool

 Jan 12, 2019

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