x and y are two real numbers, each of which is uniformly randomly chosen in the interval (1,5).
Determine the probability of each of the following events:
(i) ⌈3*x⌉ + ⌈4*y⌉ = ⌈3*x + 4*y⌉
(ii) ⌈3*x⌉ - ⌈4*y⌉ = ⌈3*x - 4*y⌉
(iii) ⌈3*x⌉ * ⌈4*y⌉ = ⌈12*x*y⌉
⌈3*x⌉
(iv) --------- = ⌈(3*x)/(4*y)⌉
⌈4*y⌉
*** ⌈n⌉ denotes ceiling(n), that is, the lowest integer greater than or equal to n.
The region in the plane looks like a bunch of right triangles. There are actually 16*12=192 shapes bounded by a hyperbola and a horizontal and vertical segment.
If we let the horizontal rank run as i from 4/15 and the vertical as j from 5 to 20. Each shape is bounded by the equations x=i/3, y=j/4, and xy=(ij-1)/12.
The area of each (i,j) shape simplifies to 1/12 - (ij-1)/12 * ln((ij-1)/(ij))
The total area is the double sum. With help from WolframAlpha
A=15.90577083
The probability sought is then (16-A)/16 = 0.005889323
A closed form is implied but in involves the sum of those 192 natural logs.
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Posted by Jer
on 2022-12-08 10:09:17 |
Part (iii)
