If a real number x is chosen at random in the interval [0,3] and a re

Question

If a real number x is chosen at random in the interval   and a real number y is chosen at random in the interval  , what is the probability that x < y?

A) 1/2 
B) 7/12 
C) 5/8 
D) 2/3 
E) 3/4

ENGRTOMBA2018 · Accepted Answer

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.

davedekoos · Answer

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  , 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

MathRevolution · Answer

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

If a real number x is chosen at random in the interval   and a real number y is chosen at random in the interval  , what is the probability that x < y?

A) 1/2 
B) 7/12 
C) 5/8 
D) 2/3 
E) 3/4

Following is my approach.... check this out...

Since x is in   and y is in   we can put them in the coordinate plane. 
That means in the coordinate plane x and y satisfy 0<= x <=3 and 0<= y <=4. 
A point (x, y) can be in the rectangle with having (0, 0), (3, 0), (0, 4), (3, 4) as its 4 vertices. 
Moreover the points with x<y should be in the trapezoid above the line through (0, 0) and (3, 3).

The area of the trapezoid above the line(joining (0, 0) and (3, 3)) is 15/2 and the area of the rectangle is 12.

So the probability is (15/2)/12 = 15/24=5/8.

The answer is, therefore, (C).

ENGRTOMBA2018 · Answer

MathRevolution , thank you. That is indeed an incorrect assumption on my end. Also, I think the text in red above is not correct. I believe you wanted to write the line joining (0,0) and (3,3) and NOT (3,4). The line joining (0,0) and (3,4) will not leave a trapezoid at the top.

bablu1234 · Answer

Thanks MathRevolution. It was tough one and I couldn't crack it.

MathRevolution · Answer

That is indeed an incorrect assumption on my end. Also, I think the text in red above is not correct. I believe you wanted to write the line joining (0,0) and (3,3) and NOT (3,4). The line joining (0,0) and (3,4) will not leave a trapezoid at the top.

Hi! Engr
Thank you for pointing out my error. As you pointed out the line is joining (0,0) and (3,3) and NOT (3,4). That makes the diagram a trapezoid.

Kinshook · Answer

Asked: If a real number x is chosen at random in the interval   and a real number y is chosen at random in the interval  , what is the probability that x < y?

Probability that x<y = (3/4)(1/2) + (1/4)(1) = 3/8 + 1/4 = (3+2)/8 = 5/8

IMO C

samarpan.g28 · Answer

bumpbot · Answer

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If a real number x is chosen at random in the interval [0,3] and a re

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