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phalanx123
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Ok here is the question
If a football match ends in a draw, there may be a ‘penalty shoot-out’. In one
version of a penalty shoot-out, each team initially takes 5 shots at the goal. If one teambscores more times than the other, then that team wins.
(i) Two teams take part in a penalty shoot-out. The probability that any player scores from a single shot is p, where 0 < p < 1. Prove carefully that the probability, [tex]\alpha[/tex], that neither side has won at the end of the initial 10-shot period is given by
[tex]\alpha=\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n}[/tex]
In the case p = 1 2 , evaluate and find the probability that the first team to shoot wins at the end of the initial period.
(ii) If neither team has won at the end of the initial period, the teams take another shot each. If one team scores and the other does not, then the team that scores wins; if neither team wins, then the teams take another shot each and the process continues until one of the teams wins. Find the probability, for 0 < p < 1, that a total of 10 + 2k + 2 shots (k >= 0) are required to win the game. Verify that the probability that, eventually, one
team or the other will win is 1.
I got fine with the first part of the question,but stuck on part (ii). Here is my attempt to the solution
I started by argueing that
the propability that 12 shots are needed is [tex]2p(1-p)\alpha[/tex]
the propability that 14 shots are needed is [tex]2ppp(1-p)+2(1-p)(1-p)p(1-p)\alpha=2p(1-p)(p^2+(1-p)^2)\alpha[/tex]
the propability that 16 shots are needed is [tex]2ppppp(1-p)+2(1-p)(1-p)(1-p)(1-p)p(1-p)\alpha=2p(1-p)(p^4+(1-p)^4)\alpha[/tex]
and so on
so the probability p of 10+2k+2 shots are needed is
[tex]p=\left\{\begin{array}{cc}2p(1-p)\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n},&\mbox{ if } k=0\\2p(1-p)(p^{2i}+(1-p)^{2i})\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n}, & \mbox{ if } k>0\end{array}\right[/tex]
where i>=1
I am not sure about this answer as it doesn't look right to me and I don't know how to verify that the probability that eventually one team or the other will win is 1 either . Could somebody help me please? Thanks a million
If a football match ends in a draw, there may be a ‘penalty shoot-out’. In one
version of a penalty shoot-out, each team initially takes 5 shots at the goal. If one teambscores more times than the other, then that team wins.
(i) Two teams take part in a penalty shoot-out. The probability that any player scores from a single shot is p, where 0 < p < 1. Prove carefully that the probability, [tex]\alpha[/tex], that neither side has won at the end of the initial 10-shot period is given by
[tex]\alpha=\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n}[/tex]
In the case p = 1 2 , evaluate and find the probability that the first team to shoot wins at the end of the initial period.
(ii) If neither team has won at the end of the initial period, the teams take another shot each. If one team scores and the other does not, then the team that scores wins; if neither team wins, then the teams take another shot each and the process continues until one of the teams wins. Find the probability, for 0 < p < 1, that a total of 10 + 2k + 2 shots (k >= 0) are required to win the game. Verify that the probability that, eventually, one
team or the other will win is 1.
I got fine with the first part of the question,but stuck on part (ii). Here is my attempt to the solution
I started by argueing that
the propability that 12 shots are needed is [tex]2p(1-p)\alpha[/tex]
the propability that 14 shots are needed is [tex]2ppp(1-p)+2(1-p)(1-p)p(1-p)\alpha=2p(1-p)(p^2+(1-p)^2)\alpha[/tex]
the propability that 16 shots are needed is [tex]2ppppp(1-p)+2(1-p)(1-p)(1-p)(1-p)p(1-p)\alpha=2p(1-p)(p^4+(1-p)^4)\alpha[/tex]
and so on
so the probability p of 10+2k+2 shots are needed is
[tex]p=\left\{\begin{array}{cc}2p(1-p)\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n},&\mbox{ if } k=0\\2p(1-p)(p^{2i}+(1-p)^{2i})\sum_{n=0}^5\left(\begin{array}{cc}5\\0\end{array}\right)^2(1-p)^{2n}p^{10-2n}, & \mbox{ if } k>0\end{array}\right[/tex]
where i>=1
I am not sure about this answer as it doesn't look right to me and I don't know how to verify that the probability that eventually one team or the other will win is 1 either . Could somebody help me please? Thanks a million