Part(a)
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_{\imath=0}^{\infty} x^{\imath}= \frac{1}{1-x} $ for |x| < 1
$ P(B) = p\sum_{\imath=0}^{\infty} (1-p)^{4\imath} = \frac{p}{1-(1-p)} $
