91Ó°ÊÓ

The probability of getting a head is\(P\left( H \right) = \frac{1}{2}\)

The probability of getting a tail is\(P\left( T \right) = \frac{1}{2}\)

Let X denote the number of tosses that are required.

The probability that at least one head and one tail have been obtained.

\(\begin{array}{l}{\rm{ = 1}} - {\rm{Pr}}\left( {{\rm{no}}\,\,{\rm{success}}} \right)\\{\rm{ = 1}} - {\rm{Pr}}\left( {no\,\,head} \right){\rm{ \times Pr}}\left( {no\,\,tail} \right)\\ = 1 - \left( {1 - \frac{1}{2}} \right) \times \left( {1 - \frac{1}{2}} \right)\\ = 1 - \left( {\frac{1}{2} \times \frac{1}{2}} \right)\end{array}\)

\(\begin{array}{l} = 1 - \frac{1}{4}\\ = \frac{3}{4}\end{array}\)

The X follows a geometric distribution with parameter p, that is \(X \sim Geo\left( {p = \frac{3}{4}} \right)\)

\(\begin{array}{c}p\left( x \right) = \left\{ \begin{array}{l}P\left( {X = x} \right) = p{\left( {1 - p} \right)^{x - 1}}\,x = 1,2, \ldots \\0,else\end{array} \right.\\\left\{ \begin{array}{l}P\left( {X = x} \right) = \frac{3}{4}{\left( {\frac{1}{4}} \right)^{x - 1}}\,x = 1,2, \ldots \\0,else\end{array} \right.\end{array}\)