91Ó°ÊÓ

(a)

Suppose that the first flip has landed on Heads. Then, if we are counting the number of flip needed to obtain both sides, we are hoping for Tail in next flips. So, if we are given that the first flip is Head, we have that the remaining number of flips that are needed to obtain both sides has Geometric distribution with parameter 1-p. Similarly, if we are given that the first flip is Tail, we have that the remaining number of flips that are needed to obtain both sides has Geometric distribution with parameter p. Therefore

• $$\begin{gathered} E\left(X\right)=E\left(X/FirstHead\right)p\left(FirstHead\right)+E\left(X/FirstTail\right)p(FirstTail) \\ =\left(1+\frac{1}{1-p}\right)p+\left(1+\frac{1}{p}\right)\left(1-p\right)=1+\frac{p^2+{\left(1-p\right)}^2}{p\left(1-p\right)} \end{gathered}$$
  

(b)

No matter what is the number of total flips, the last flip will be Head with probability $p$and Tail with probability 1-p. This is because of the fact that all flips are independent and equally distributed.

(a) Condition the expectation on the first flip.

(b) The probability is p.