Originally posted by davegage
Suppose two co-workers each take one lunch break per day. They arrive at the cafeteria independently at random times between 12 PM and 1 PM, and each stays for exactly x minutes. The probability that either co-worker arrives while the other is in the cafeteria is 40%.
If x = a - b*(c)^0.5, then find (a + b + c).
You also are given: a, b, and c a ...[text shortened]... of you can solve it to either confirm my answer or show me why my answer is wrong.
Let's see what we get:
Let A be the arrival time of one co-worker, and B the arrival time of the other. The the probability that the second worker arrives while the first one is there is P(0 <= (B-A) <= x). Using independence, this is integral(f(t)P(t <= B <= t+x))dt, where f is the pdf of A. At this point we have to assume 'at random' means 'uniformly at random' (why do pre-university examiners always pretend that random means uniform?), so f(t) is 1/60 over the range 0 to 60 (I'm calling 12 PM 0 for convenience), and the integral becomes
integral[0,60-x](1/60 * x/60)dt + integral[60-x,60](1/60*(60-t)/60)dt
= (60-x)x/3600 + (t/60 -(t^2)/7200)[60-x,60]
= (60-x)x/3600 + 1/2 - (60-x)/60 + (60-x)(60-x)/7200
= (60-x)(60+x)/7200 - (30-x)/60
which is 0.4. For convenience let x = 60k:
(1-k)(1+k)/2 - (1-2k)/2 = 0.4
1 - k^2 - 1 + 2k - 0.8 = 0
k^2 -2k +0.8 = 0
=> k = (2 +-sqrt(4-3.2))/2
= 1 +-sqrt(0.8)/2
so x = 60 +-30*sqrt(0.8)
= 60 +-6*sqrt(20)
Now why does this specify a,b and c uniquely? The answer is that the roots of positive squarefree integers are linearly independent over the rationals, but I don't want to prove that here. For your purposes I think you're just allowed to assume that the solution is unique.
So a = 60, b = 6, c = 20 and their sum is 86.
Did I get the same as you?