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

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New post 13 Dec 2015, 19:29
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A
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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
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If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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New post Updated on: 14 Dec 2015, 07:57
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bablu1234 wrote:
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


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

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New post Updated on: 21 Dec 2015, 01:12
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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 [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



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

Since x is in [0,3] and y is in [0,4] 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).
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Originally posted by MathRevolution on 14 Dec 2015, 01:17.
Last edited by MathRevolution on 21 Dec 2015, 01:12, edited 2 times in total.
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Re: If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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New post 14 Dec 2015, 07:46
MathRevolution wrote:
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 [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

Hi! Engr
I'm afraid that you made a mistake to understand the problem because x and y are not integers but real numbers.
So we cannot count the all the cases of (x, y) satisfying x<y.

The following is my approach.... check this out...

Since x is in [0,3] and y is in [0,4] 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, 4).

The area of the trapezoid above the line(joining (0, 0) and (3, 4)) 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).



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.
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Re: If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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New post 14 Dec 2015, 11:54
MathRevolution wrote:
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 [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

Hi! Engr
I'm afraid that you made a mistake to understand the problem because x and y are not integers but real numbers.
So we cannot count the all the cases of (x, y) satisfying x<y.

The following is my approach.... check this out...

Since x is in [0,3] and y is in [0,4] 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, 4).

The area of the trapezoid above the line(joining (0, 0) and (3, 4)) 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).


Thanks MathRevolution. It was tough one and I couldn't crack it.
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If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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New post 17 Dec 2015, 12:53
1
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
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Re: If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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New post 21 Dec 2015, 01:20
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.
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Re: If a real number x is chosen at random in the interval [0,3] and a re  [#permalink]

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