A) \[2:1\]
B) \[3:2\]
C) \[1:2\]
D) \[2:3\]
Correct Answer: C
Solution :
| Let A = Event of A's win |
| B = Event of B riding A |
| C = Event of C riding A |
| Then \[\operatorname{P}\left( \frac{A}{B} \right)=\frac{1}{6}and\,P\left( \frac{A}{C} \right)=3\left( \frac{1}{6} \right)=\frac{1}{2}\] |
| Now \[\operatorname{A}=A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right)\] |
| Since B and C are mutually exclusive we have |
| \[\operatorname{P}\left( A \right)=P\left( \operatorname{A}\cap \operatorname{B} \right)+\operatorname{P}\left( \operatorname{A}\cap \operatorname{C} \right)\] |
| \[=\operatorname{P}\left( B \right)P\left( \frac{A}{B} \right)+\operatorname{P}\left( \operatorname{C} \right)\operatorname{P}\left( A/C \right)\] |
| \[\frac{1}{2}\times \frac{1}{6}+\frac{1}{2}\times \frac{1}{2}=\frac{1+3}{12}=\frac{1}{3}\] |
| Therefore \[\operatorname{P}\left( \overline{A} \right)=1-\frac{1}{3}=\frac{2}{3}\] |
| Hence odds in favour is \[\operatorname{P}\left( E \right):P\left( \overline{\operatorname{E}} \right)=1:2\] |
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