Thanks for all the help. I have one more question I hope I can get some pointers on.

Question: Three random variables X1, X2 and X3 are generated by spinning a continuously calibrated (from 0 to 1) fair wheel three times (see 2nd attached image). The random variables X1, X2 and X3 are independent and uniformly distributed on [0,1]. Find the probability that none of the values X1, X2, X3 are within +or- d of each other by using order statistics and the following steps:

a)Let 0<=Y1<=2<=Y3<=1 be the order statistics associated with the random variables X1, X2, and X3. Since we are only interested in the relative distance between Y1, Y2 and Y3 we do note change the results by always referencing them to the value of the minimum Y1 and simple considering the joint distribution fY2Y3(y2,y3)=2!fx(Y)*fx(y)=

b)Now show that the probability for Y1,Y2 and Y3 to be within +or- d of each other is given by

*see 1st image attached to this message

Show this by describing where y2 and then y3 should be placed on the interval [0,1] in order to satisfy the +or-d separation conditions on the locations of Y2, and Y3 relative to each other and to Y1 (at the origin)

c) Place the appropriate limits on the double integral and integrate the expression to obtain the result. Set d=0.1 and show that you obtain a probability that all three are within +or- 0.1 is 0.49 or 49%

d)Give a geometrical derivation of the result of part c by using the limits to the integrals found above to draw the region of interest in the y2,y3 plane and finding the probability as the ratio of this area to the total area for which Y3>Y2.

Thank you for the help!