1b) Once again the
sample space S is {HH, HT, TH, TT}. Let E be the event of obtaining one head and one tail,
so E = {HT, TH}.
Hence P(E) = n(E) / n(S) = 2/4 =1/2.
Since each toss is independent of all previous tosses the
probability of obtaining a mixture of a head and a tail five times in a row is
1/2 ´ 1/2 ´ 1/2 ´ 1/2 ´ 1/2 = 1/32.