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 **
**m****s)/10**^{6}**
= 0.000**12

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 =
S _{1}^{432000 }Po(n) x
p(n) = 0.**2457

** **

The chance of getting a
collision failure is **~25%** in the mentioned 12 hour test.