question_answer

In a hockey match, both teams A and B scored same number of goals upto the and of the game, so to decide the winner, the referee asked both the captains to throw a die alternatively and decided that the team, whose captain gets a six first, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match.

Answer:

Let \[{{E}_{1}}=\] Event of A getting six and \[{{E}_{2}}=\] Event of B getting six In throwing a die, Total number of outcomes, n(S) = 6 \[\therefore \]      \[P({{E}_{1}})=P({{E}_{2}})=\frac{1}{6}\] Then,    \[P({{\bar{E}}_{1}})=P({{\bar{E}}_{2}})=1-\frac{1}{6}=\frac{5}{6}\] Since, the referee gives first chance to captain A for throwing a die. \[\therefore \] Probability of A winning, \[P(A)=[P({{E}_{1}})+P({{\bar{E}}_{1}}\cap {{\bar{E}}_{2}}\cap {{E}_{1}})+....]\]             \[=P({{E}_{1}})+P({{\bar{E}}_{1}})\cdot P({{\bar{E}}_{2}})\cdot P({{E}_{1}})+....\]             \[=\frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+...\]             \[=\frac{1}{6}\left[ 1+{{\left( \frac{5}{6} \right)}^{2}}+... \right]=\frac{1}{6}\left( \frac{1}{1-\frac{25}{36}} \right)\] \[\left[ \begin{align}   & \because \,\,\text{sum}\,\,\text{of}\,\,\text{an}\,\,\text{infinite}\,\,GP=\frac{a}{1-r} \\  & \text{and}\,\,\text{here,}\,\,a=1,\,\,r=\left( \frac{{{5}^{2}}}{6} \right)=\frac{25}{36} \\ \end{align} \right]\] \[=\frac{1}{6}\times \frac{36}{11}=\frac{6}{11}\] and probability of B winning, \[P(B)=P({{\bar{E}}_{1}}\cap {{E}_{2}})+P({{\bar{E}}_{1}}\cap {{\bar{E}}_{2}}\cap {{\bar{E}}_{1}}\cap {{E}_{2}})+...\] \[=P({{\bar{E}}_{1}})\cdot P({{E}_{2}})+P({{\bar{E}}_{1}})\cdot P({{\bar{E}}_{2}})\cdot P({{\bar{E}}_{1}})\cdot P({{E}_{2}})+...\] \[=\frac{5}{6}\times \frac{1}{6}+\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}+...\] \[=\frac{5}{36}\left[ 1+{{\left( \frac{5}{6} \right)}^{2}}+... \right]=\frac{5}{36}\left( \frac{1}{1-\frac{25}{36}} \right)\] \[\left[ \because \,\,\,\text{sum}\,\,\text{of}\,\,\text{an}\,\,\text{infinite}\,\,GP=\frac{a}{1-r} \right]\] \[=\frac{5}{36}\times \frac{36}{11}=\frac{5}{11}\]