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.
My take on part 3.
First I made a substitute u=3x and v=3y. Then u is on (3,15) and v is on (4,20).
Then (iii) becomes ceil[u] * ceil[v] = ceil[u*v]
Then I split u and v into a integer part and fractional part.
p=integer 4 to 15
q=integer 5 to 20
r=fraction 0 to 1
s=fraction 0 to 1
Then u=p-r and v=q-s.
Substituting these in gives ceil[p-r] * ceil[q-s] = ceil[(p-r)*(q-s)]
The domain can be broken up into 12*16=192 unit squares were each of p and q are fixed for that square
Now I'm going to focus on a single one of those unit squares:
ceil[p-r] * ceil[q-s] = ceil[(p-r)*(q-s)]
ceil[p-r] * ceil[q-s] = ceil[pq - ps - qr - rs]
Pull the integers out:
(p + ceil[r]) * (q + ceil[s]) = pq + ceil[-ps - qr + rs]
The ceil[] on the left resolve to zero, so the left side collapses to pq.
Then 0 = ceil[-ps - qr + rs].
This exactly equals 0 when r and s are zero. Then the area within the unit square we want is where -ps-qr+rs > -1.
This is a branch of a hyperbola, as Jer pointed out. It has an r-intercept of (1/q, 0) and an s-intercept (0, 1/p)
If I solve for s: s = (rq-1)/(r-p). I can integrate this over r from 0 to 1/q to get the area.
After a bit of tedious calculations I get an area of 1+(pq-1)*ln((pq-1)/pq). This is for a single one of the unit squares. Then the total probability is given by adding up and averaging all 192 squares, which is the double sum (1/192) * sigma {p=4 to 15} sigma {q=5 to 20} ( 1+(pq-1)*ln(1-1/pq) ).
I pulled out a spreadsheet and after a bit of playing with an auto-fillable formula I calculated all 192 values and took the average to get a final result of 0.00588932. (as far as the precision of the spreadsheet will get)
Part 3
