Random + Random + Random (Posted on 2009-08-28)

	Submitted by Jer
Rating: 3.0000 (1 votes)
Solution:	(Hide)
This can be pictured as the the volumes of regions within a unit cube. The pyramid with corners (1,1,0), (1,0,1), (0,1,1), and (1,1,1) where 2 < A+B+C < 3 so X=2 the volume here is 1/6. For X+9 the total volume is 1/6(1-.9^3)+ 1/6(.1^3-.09^3)+1/6(.01^3+.009^3)... =1/6(.271 + .000271 + .000000271...) =1/6(.271271271... =1/6(271/999) =271/5994 Similar reasoning can be used to derive the other values of X. Remembering to add the 2/3 for 1 and 1/6 for 2 yields the following distribution: P(X)=1 = 4003/5994 P(X)=2 = 1018/5994 P(X)=3 = 37/5994 P(X)=4 = 61/5994 P(X)=5 = 91/5994 P(X)=6 = 127/5994 P(X)=7 = 169/5994 P(X)=8 = 217/5994 P(X)=9 = 271/5994

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2024-03-12 08:11:55
	solution	Charlie	2009-08-28 17:59:35