A) \[5:13\]
B) \[5:18\]
C) \[13:5\]
D) None of these
Correct Answer: C
Solution :
| [c] Let E= the event that horse A wins \[{{E}_{1}}=\]the event that jockey B rides hors A \[{{E}_{2}}=\]the event that jockey C rides hors A According to question odds in favour of \[{{E}_{1}}=2:1\] |
| \[\therefore P({{E}_{1}})=\frac{2}{3}\] |
| and \[P\left( \frac{E}{{{E}_{1}}} \right)=\frac{1}{6}\] (Since, when B rides A, all six |
| Horses are equally likely to win) |
| \[P({{E}_{2}})=1-P({{E}_{1}})=1-\frac{2}{3}=\frac{1}{3}\] |
| and \[P\left( \frac{E}{{{E}_{2}}} \right)=3P\left( \frac{E}{{{E}_{1}}} \right)=\frac{1}{2}\] |
| Let \[{{A}_{1}}={{E}_{1}}\cap E\] and \[{{A}_{2}}={{E}_{2}}\cap E\] |
| Now, required probability |
| \[P(E)=P({{A}_{1}})+P({{A}_{2}})\] |
| \[=p({{E}_{1}}\cap E)+P({{E}_{1}}\cap E)\] |
| \[=p({{E}_{1}})P\left( \frac{E}{{{E}_{1}}} \right)+P({{E}_{2}})P\left( \frac{E}{{{E}_{2}}} \right)\] |
| \[=\frac{2}{3}.\frac{1}{6}+\frac{1}{3}.\frac{1}{2}=\frac{5}{18}.\] |
| So, that odds against winning of A are 13 : 5. |
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