The DF-11 squitter has a 60 ms duration and happens once a second. Since the UF-0 interrogation has 30 ms, the chance of any part of an interrogation happening while squittering is:
1) p = (60+2x30 ms)/106 = 0.00012
There is a 12 hour test that fails when a collision is detected. For this test, 10 interrogations are made each second. Events with such a low probability as Squitter-Interrogation collisions follow the Poison probability distribution .
2) p(n/N) = (Np)n e-Np/n!
Where N is
the number of interrogations in 12 hours (432000) and n is the number of
collisions. P(n/N) is the probability of having precisely n collisions in N
tries (interrogations).
Equation 2 is the limit of the binomial distribution for N -> ∞, while the product Np stays finite… or in practical terms Np << N or p << 1 (rare events). In our case this condition is very much true:
3) Np = 432000*0.00012 = 51.84
So, our Poison distribution will be:
4) p(n) = (51.84)n * 3.06x10-23 /n!
But not all interrogations are observed, of the 432000 interrogations, only 2356 will be sampled by the computer (1 sample every 5 seconds), the probability of not observing a single collision in any of those samples, if n of them will happen is:
5) Qo(n) = (1-n/432000)2356
Then the probability of observing one or more would be:
6) Po(n) = 1 – Qo(n)
The probability of observing a collision when precisely n of them happen, will be the probability of observing Po multiplied by the probability P(n) of having precisely n collisions. The probability of observing a collision, disregarding how many actually happened, would be the sum of the probabilities for each n form 1 to (in principle) 432000:
6) P = S1432000 Po(n) x p(n) = 0.2457
The chance of getting a collision failure is ~25% in the mentioned 12 hour test.