Show that P(B) > P(C) > P(T) > P(A):
- P(H) = p , 0 < p < 1
$ P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} $
Recall geometric series:
$ \sum_{i=0}^\infty x^i = 1\{1-x} $ , for |x| < 1
