Question

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

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?

Here's what I got - I'm hoping someone can tell me if this is a valid argument.

P(x<y) = P(y>3) + P(y<3 and x<y|y<3) = 1/4 + (3/4)(1/2) = 5/8.

I know 5/8 is the correct answer. Is this the correct method to get there? Thanks.

Math   Statistics-And-Probability   
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Homework Answers

Answer #1

Yes, the solution given by you is correct. It is a valid argument.

Following the another method that can be used:

Refer to the attached picture for the figure. We are given that 0 ≤ x ≤ 3 and 0 ≤ y ≤ 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 units2

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.

The correct answer is 5/8

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