A) \[9/10\]
B) \[9/11\]
C) \[10/11\]
D) \[7/11\]
Correct Answer: B
Solution :
| [b] Let |
| Event \[{{E}_{1}}:\] A wins the match at the end of 11th game |
| Event \[{{E}_{2}}:\] B wins exactly one game |
| A can win the match at the end of 11th game by winning exactly one of the first 10 games and drawing the remaining 9. |
| (ii) or by winning exactly one of the first 10 games, losing exactly one and drawing the remaining 8. |
| \[\therefore P({{E}_{1}}){{=}^{10}}{{C}_{1}}\left( \frac{1}{2} \right){{\left( \frac{1}{3} \right)}^{9}}\times \frac{1}{2}{{+}^{10}}{{C}_{2}}\left( \frac{1}{2} \right){{\left( \frac{1}{3} \right)}^{8}}\left( \frac{1}{6} \right)\left( \frac{1}{2} \right)(2!)\]\[\therefore \,\,\,P({{E}_{2}}/{{E}_{1}})=\frac{^{10}{{C}_{2}}\left( \frac{1}{2} \right){{\left( \frac{1}{3} \right)}^{8}}\left( \frac{1}{6} \right)(2!)}{^{10}{{C}_{1}}{{\left( \frac{1}{2} \right)}^{2}}{{\left( \frac{1}{3} \right)}^{9}}{{+}^{10}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{2}}{{\left( \frac{1}{3} \right)}^{8}}\left( \frac{1}{6} \right)(2!)}=\frac{9}{11}\] |
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