Two real numbers x and y are randomly chosen from the interval (0,1). If we know that lx-yl <1/2, what is the probability that x<1/2<y

Answer 1

Given that |x-y|<1/2 we no that the square 0 to 1 in x and 0 to 1 in y are all possible (x,y). The restriction defines an area left when you take off the corner from (0,1/2) to (1/2,1) and the opposite corner from (1/2,0) to (1,1/2) This comes from solving the inequalities x - y < 1/2 for y <x and x-y > -1/2 for y >x (the equalities give you the two lines that cut off the corners. The area left in the center is 1 - 2(1/2(1/2 x 1/2))) = 3/4

Now the probability that x < 1/2 and 1/2 < y is also the same 1/8 area triangle (It completes the upper left quater of the square from the omitted values from the first condition)

The probability is the area of the second condition and allowed by the first condition divided by the central region allowed by the 1st condition or 1/8/(3/4) = 1/6

Answer 2

Hi Alisa T.,

You can build your answer in the following way. Pick your variable values one at a time, and examine!

(1) The odds of x<1/2 are 0.5

(2) Given an x<1/2, what are the odds of y>1/2? Since they are independent, also 1/2. That gives you P=1/4, so far.

(3) The tricky part is the absolute value requirement. But if you let x range from 0 to 1/2 in value, and then consider the odds (probability) of y being requisitely close, you can see that that probability ranges from 0 (for x~0) to 1 (for x~1/2). The actual endpoints don't count in the calculation, they're only single points out of an infinity of choices! Therefore the average probability of it (y) being close enough is just the average of those two probabilities, namely 0.5. Therefore, the overall P=1/8.

So, what did you learn about multiple constraint problems? Hopefully, that you can attack them one bit at a time, and use successive conditional probabilites to calculate. Occasionally, you may have to trot out Bayes' Theorem, or some such, but that's another whole story!

-- Cheers, --Mr. d.