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![If a real number x is chosen at random in the interval [0,3] and a re](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If a real number x is chosen at random in the interval [0,3] and a re"){kind=icon}
13 Dec 2015, 19:29
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (02:00) correct
65%
(01:57)
wrong
based on 277
sessions
History
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Time
Result
Not Attempted Yet
If a real number x is chosen at random in the interval [0,3] and a real number y is chosen at random in the interval [0,4], what is the probability that x < y?
A) 1/2
B) 7/12
C) 5/8
D) 2/3
E) 3/4
A) 1/2
B) 7/12
C) 5/8
D) 2/3
E) 3/4
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,315
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
Updated on: 14 Dec 2015, 07:57
Originally posted by ENGRTOMBA2018 on 13 Dec 2015, 20:14.
Last edited by ENGRTOMBA2018 on 14 Dec 2015, 07:57, edited 2 times in total.
Last edited by ENGRTOMBA2018 on 14 Dec 2015, 07:57, edited 2 times in total.
Updated the solution
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bablu1234
Refer to the attached picture for the figure. We are given that \(0 \leq x \leq 3\) and \(0 \leq y \leq 4\)
Thus the region x-y when plotted gives you a rectangle with base = 3 units and height 4 units (as shown in the picture), giving you a total area = 4*3=12 \(units^2\)
Now, realize that y=x is a line that passes through (0,0) and (3,3) and divides the above rectangle into a trapezoid (ABCD) and a triangle. The area y>x will belong to the trapezoidal area.
Thus, the area of the trapezoid = 0.5*(4+1)*3 = 15/2
Finally, the required probability = trapezoid area / total area = (15/2)/12 = 12/24 = 5/8.
C is the correct answer.
Hope this helps.
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12-14-15 9-53-57 AM.jpg [ 26.2 KiB | Viewed 34312 times ]
17 Dec 2015, 12:53
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Another way to look at the problem:
Break up the range of y into two parts.
1. \(y>3\)
2. \(0\leq{y}\leq{3}\)
We can consider the probability of \(y>x\) for each part, then add them up.
For part 1, if \(y>3\), then it will always be greater than x no matter what x is. y will be in this range 1/4 of the time. Probability = 1*1/4 = 1/4.
For part 2, both x and y are in the range of [0,3], so we can logically conclude that \(y>x\) half the time (similarly \(x>y\) half the time). y will be in this range 3/4 of the time. 1/2*3/4 = 3/8
Add up both parts:
1/4 + 3/8 = 5/8
Answer C
Break up the range of y into two parts.
1. \(y>3\)
2. \(0\leq{y}\leq{3}\)
We can consider the probability of \(y>x\) for each part, then add them up.
For part 1, if \(y>3\), then it will always be greater than x no matter what x is. y will be in this range 1/4 of the time. Probability = 1*1/4 = 1/4.
For part 2, both x and y are in the range of [0,3], so we can logically conclude that \(y>x\) half the time (similarly \(x>y\) half the time). y will be in this range 3/4 of the time. 1/2*3/4 = 3/8
Add up both parts:
1/4 + 3/8 = 5/8
Answer C
