BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

probability problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 66
Joined: Wed Feb 20, 2008 6:23 pm
Thanked: 2 times

probability problem

by anjaligeorge1 » Fri Jun 06, 2008 11:22 am
In the xy-plane, a triangle have vertex (0,0), (4,0) and (4,5). If point (x,y) is selected at random from the triangular region, what is the probability that x-y>0?

(A) 1/5
(B) 1/3
(C) 1/2
(D) 2/3
(E) 4/5
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 178
Joined: Wed May 14, 2008 3:51 pm
Thanked: 16 times
Followed by:1 members
GMAT Score:99%+

by egybs » Fri Jun 06, 2008 1:26 pm
We know that the triangle has a base of 4 and a height of 5 which has an area of 10.

The area within the triangle where x-y>0 is bound by the lines y=x, y=0 and x=4. You can quickly see that this new, smaller triangle has a base of length 4 and a height of length 4... therefore the area is 8.

So, there is a 8/10 chance that the point will be x-y>0. Or 4/5. OA is E?
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Mon Apr 28, 2008 5:33 am
Thanked: 2 times

by ramyaravindran » Fri Jun 06, 2008 10:26 pm
x can be any of the values 0,1,2,3,4
y can be any of the values 0,1,2,3,4,5

So the possible combinations where x-y > 0 are (1,0), (2,0), (2,1),(3,0),(3,1),(3,2), (4,0),(4,1),(4,2),(4,3) ==> 10 possible combinations out of a total of 30 combinations. So the probability is 10/30 or 1/3.

Is the answer A?
Top
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Mon Apr 28, 2008 5:33 am
Thanked: 2 times

by ramyaravindran » Fri Jun 06, 2008 10:28 pm
Correction to my previous post - Is the answer B?
Top
Master | Next Rank: 500 Posts
Posts: 178
Joined: Wed May 14, 2008 3:51 pm
Thanked: 16 times
Followed by:1 members
GMAT Score:99%+

by egybs » Fri Jun 06, 2008 11:56 pm
This is incorrect for a few reasons... among them:
1) no reason x and y can only be integers... for example, (4,3.999999999999999) is perfectly fine point.
2) your logic on x being anything between 0 and 4 and y being anything between 0 and 5 is flawed. For example, the point (0,5) does not exist within the triangle... but you have counted it. What you have described is a rectangle.

The best approach is to consider the area of the two triangles described. See my description.
ramyaravindran wrote:x can be any of the values 0,1,2,3,4
y can be any of the values 0,1,2,3,4,5

So the possible combinations where x-y > 0 are (1,0), (2,0), (2,1),(3,0),(3,1),(3,2), (4,0),(4,1),(4,2),(4,3) ==> 10 possible combinations out of a total of 30 combinations. So the probability is 10/30 or 1/3.

Is the answer A?
Top
User avatar
Legendary Member
Posts: 986
Joined: Wed Dec 20, 2006 11:07 am
Location: India
Thanked: 51 times
Followed by:1 members

by gabriel » Sat Jun 07, 2008 4:11 am
egybs wrote:We know that the triangle has a base of 4 and a height of 5 which has an area of 10.

The area within the triangle where x-y>0 is bound by the lines y=x, y=0 and x=4. You can quickly see that this new, smaller triangle has a base of length 4 and a height of length 4... therefore the area is 8.

So, there is a 8/10 chance that the point will be x-y>0. Or 4/5. OA is E?
This is exactly how I solved it, my answer is 4/5 too.
Top
Senior | Next Rank: 100 Posts
Posts: 66
Joined: Wed Feb 20, 2008 6:23 pm
Thanked: 2 times

by anjaligeorge1 » Sat Jun 07, 2008 5:57 am
The answer is 4/5
Top
Post new topic Post Reply
• Page 1 of 1