1994 Paper 3 Q12

Year: 1994
Paper: 3
Question Number: 12

Course: LFM Stats And Pure
Section: Independent Events

Difficulty: 1700.0 Banger: 1473.3

probability independent events geometric series tennis conditional probability algebraic manipulation infinite series

Problem

In certain forms of Tennis two players \(A\) and \(B\) serve alternate games. Player \(A\) has probability \(p\low_{A}\) of winning a game in which she serves and \(p\low_{B}\) of winning a game in which player \(B\) serves. Player \(B\) has probability \(q\low_{B}=1-p\low_{B}\) of winning a game in which she serves and probability \(q\low_{A}=1-p\low_{A}\) of winning a game in which player \(A\) serves. In Shortened Tennis the first player to lead by 2 games wins the match. Find the probability \(P_{\text{short}}\) that \(A\) wins a Shortened Tennis match in which she serves first and show that it is the same as if \(B\) serves first. In Standard Tennis the first player to lead by 2 or more games after 4 or more games have been played wins the match. Show that the probability that the match is decided in 4 games is \[ p^{2}_Ap_{B}^{2}+q_{A}^{2}q_{B}^{2}+2(p\low_{A}p\low_{B}+q\low_{A}q\low_{B})(p\low_{A}q\low_{B}+q\low_{A}p\low_{B}). \] If \(p\low_{A}=p\low_{B}=p\) and \(q\low_{A}=q\low_{B}=q,\) find the probability \(P_{\text{stan}}\) that \(A\) wins a Standard Tennis match in which she serves first. Show that \[ P_{\text{stan}}-P_{\text{short}}=\frac{p^{2}q^{2}(p-q)}{p^{2}+q^{2}}. \]

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1473.3

Banger Comparisons: 2

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Problem source

In certain forms of Tennis two players $A$ and $B$ serve alternate
games. Player $A$ has probability $p\low_{A}$ of winning a game in which
she serves and $p\low_{B}$ of winning a game in which player $B$ serves.
Player $B$ has probability $q\low_{B}=1-p\low_{B}$ of winning a game in
which she serves and probability $q\low_{A}=1-p\low_{A}$ of winning a game
in which player $A$ serves. In Shortened Tennis the first player
to lead by 2 games wins the match. Find the probability $P_{\text{short}}$
that $A$ wins a Shortened Tennis match in which she serves first
and show that it is the same as if $B$ serves first. 

In Standard Tennis the first player to lead by 2 or more games after
4 or more games have been played wins the match. Show that the probability
that the match is decided in 4 games is 
\[
p^{2}_Ap_{B}^{2}+q_{A}^{2}q_{B}^{2}+2(p\low_{A}p\low_{B}+q\low_{A}q\low_{B})(p\low_{A}q\low_{B}+q\low_{A}p\low_{B}).
\]
If $p\low_{A}=p\low_{B}=p$ and $q\low_{A}=q\low_{B}=q,$ find the probability $P_{\text{stan}}$
that $A$ wins a Standard Tennis match in which she serves first.
Show that 
\[
P_{\text{stan}}-P_{\text{short}}=\frac{p^{2}q^{2}(p-q)}{p^{2}+q^{2}}.
\]

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