91影视

Note that the chess game in the given case cannot end up as a draw (given in the exercise).

The player who wins in the last game played (\({x^{th}}\)game) has to be the winner (\({10^{th}}\)win) because otherwise they would keep playing. For\(x \in \{ 10,11,...,19\} \)the following is true 鈥�

\({\rm{P(X = x) = P}}\left( {\left\{ {{\rm{A wins in }}{{\rm{x}}^{{\rm{th}}}}{\rm{ game }}} \right\} \cup \left\{ {{\rm{B wins in }}{{\rm{x}}^{{\rm{th}}}}{\rm{ game }}} \right\}} \right)\)

\((1)\): player\(A\)is the winner or player\(B\)is the winner.

(a)

The event\({\rm{\{ A wins in }}{{\rm{x}}^{{\rm{th}}}}{\rm{ game\} }}\)can be written as 鈥�

\({{\rm{A}}_{\rm{1}}}{\rm{ = }}\left\{ {{\rm{player}}} \right.{\rm{A wins 9 games in x - 1 games and wins }}{{\rm{x}}^{{\rm{th }}}}{\rm{game\} }}\)

Find this probability by the multiplication rule (the successive games are independent), therefore 鈥�

\(\begin{aligned}{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right) &= P( player A wins 9 games in x - 1 games) \cdot {\rm{P(S)}}\\\ &= \left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right){{\rm{p}}^{\rm{9}}}{{\rm{(1 - p)}}^{{\rm{x - 10}}}} \cdot {\rm{p}}\\ &= \left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right){{\rm{p}}^{{\rm{10}}}}{{\rm{(1 - p)}}^{{\rm{x - 10}}}}\end{aligned}\)

There are\(\left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right)\)ways to take\({\rm{9}}\)elements out of\({\rm{x - 1}}\). The\({{\rm{p}}^9}\)represents the probability of the\({\rm{9S's}}\)and\({{\rm{(1 - p)}}^{{\rm{x - 10}}}}\)represent the probability of the\({\rm{(x - 1 - 9) F's}}\).

The same way we compute the probability for the event 鈥�

\({{\rm{B}}_{\rm{1}}}{\rm{ = }}\left\{ {{\rm{player}}} \right.{\rm{ B wins 9 games in x - 1 games and wins }}{{\rm{x}}^{{\rm{th }}}}{\rm{game\} }}\)

Therefore, for\(x \in \{ 10,11,...,19\} \)the following is true 鈥�

\(\begin{aligned}P(X = x) &= P \left( {{{\rm{A}}_{\rm{1}}} \cap {{\rm{B}}_{\rm{1}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right)\\&= \left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right){{\rm{p}}^{{\rm{10}}}}{{\rm{(1 - p)}}^{{\rm{x - 10}}}}{\rm{ + }}\left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right){{\rm{p}}^{{\rm{x - 10}}}}{{\rm{(1 - p)}}^{{\rm{10}}}}\\ & = \left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right)\left( {{{\rm{p}}^{{\rm{10}}}}{{{\rm{(1 - p)}}}^{{\rm{x - 10}}}}{\rm{ + }}{{\rm{p}}^{{\rm{x - 10}}}}{{{\rm{(1 - p)}}}^{{\rm{10}}}}} \right)\end{aligned}\)

Therefore, the expression is \(\left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right)\left( {{{\rm{p}}^{{\rm{10}}}}{{{\rm{(1 - p)}}}^{{\rm{x - 10}}}}{\rm{ + }}{{\rm{p}}^{{\rm{x - 10}}}}{{{\rm{(1 - p)}}}^{{\rm{10}}}}} \right)\).

(b)

In this case, the draw is possible and the probability is given. The possible values of \({\rm{X}}\) are all positive integers bigger than \(10\) (they can play draw endlessly). Therefore, for \(x \in \{ 10,11,...,19\} \), similarly as in (a), just with different probabilities, it is 鈥�

\({\rm{P(X = x) = }}\left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right)\left( {{{\rm{p}}^{{\rm{10}}}}{{{\rm{(1 - p)}}}^{{\rm{x - 10}}}}{\rm{ + (1 - q}}{{\rm{)}}^{{\rm{x - 10}}}}{{\rm{q}}^{{\rm{10}}}}} \right)\)

Note that the draw changes only the probabilities and the fact that \({\rm{X}}\) can take values \(x \ge 20\).

Finally, the following holds 鈥�

\(\begin{aligned}{\rm{P(X}} \ge {\rm{20) = 1 - P(X < 20) = }}\sum\limits_{{\rm{x = 10}}}^{{\rm{19}}} {\rm{P}} {\rm{(X = x)}}\\{\rm{ = }}\sum\limits_{{\rm{x = 10}}}^{{\rm{19}}} {\left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right)} \left( {{{\rm{p}}^{{\rm{10}}}}{{{\rm{(1 - p)}}}^{{\rm{x - 10}}}}{\rm{ + (1 - q}}{{\rm{)}}^{{\rm{x - 10}}}}{{\rm{q}}^{{\rm{10}}}}} \right)\end{aligned}\)

Therefore, the expression is \(\sum\limits_{{\rm{x = 10}}}^{{\rm{19}}} {\left( {\begin{array}{*{20}{c}}{{\rm{x - 1}}}\\{\rm{9}}\end{array}} \right)} \left( {{{\rm{p}}^{{\rm{10}}}}{{{\rm{(1 - p)}}}^{{\rm{x - 10}}}}{\rm{ + (1 - q}}{{\rm{)}}^{{\rm{x - 10}}}}{{\rm{q}}^{{\rm{10}}}}} \right)\).