Assume a binomial distribution for probability.

We have: (p+(1-p))¹⁵=1 where p=20% or 0.2 (success) and 1-p=80% or 0.8 (failure).

When we expand the binomial we get (1-p)¹⁵+15p(1-p)¹⁴+15*14/2*p²(1-p)¹³+15*14*13/6*p³(1-p)¹²+...

In words, each term means: probability of no “live calls”, probability of just one (15 ways of arranging one in 15), probability of 2 (15*14/2=105) placed within the 15 calls, probability of 3 within the 15 (15*14*13/6=455), ...

We want this fourth term: 455(0.2³)(0.8¹²)=0.25 or 25% approximately, for exactly 3 successful live calls.

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