What is the probability that there will be 93.75 percent or more HEADS in n tosses?

Answer 1

For n in [1, 16], Prob = nC0*2-n.

For n in [17, 32] Prob = (nC0+nC1)*2-n

For n in [33, 48] Prob = (nC0+nC1+nC2)*2-n

In general,

For n in [(k-1)*16, k*16] Prob = (nC0+nC1+nC2+...+nCk-1)*2-n.