Assuming that the Poisson distribution is applicable:
The Poisson distribution is f(k) = λ^k * e^(-λ) / k!
Where:λ is the expected number of circumstances of the occurrence during interval t ;
In the problem, r = 10 calls/hour. Where it is also = 1/360 calls/sec , therefore λ = t/360 calls during interval t ;
f(k) is the probability of exactly "k" occurrences;
So based from the problem: f = 0.9 ; k = 0 ; λ = t/360
0.9 = (t/360)^0 * e^(-t/360) / 0!
0.9 = e^(-t/360)
t = -360 ln 0.9 = 37.93 s
it is approximated: 38 s is the interval with 0.9 probability of no calls received
