Question

The odds against a certain event are 5 to 2 and the odds in favour of another event, independent to the former are 6 to 5. Find the probability that (i) at least one of the events will occur, and (ii) none of the events will occur.

Answer 1

\[\text{ The odds against event A are 5 to 2. } \]
\[P\left( A \right) = \frac{2}{5 + 2} = \frac{2}{7}\]
\[\text{ The oddsin favour of event B are 6 to 5 } .\]
\[P\left( B \right) = \frac{6}{6 + 5} = \frac{6}{11}\]
\[\left( i \right) P\left( \text{ atleast one event occurs } \right)\]
\[ = P\left( A \cup B \right)\]
\[ = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\]
\[ = P\left( A \right) + P\left( B \right) - P\left( A \right) \times P\left( B \right)\]
\[ = \frac{2}{7} + \frac{6}{11} - \frac{2}{7} \times \frac{6}{11}\]
\[ = \frac{22 + 42}{77} - \frac{12}{77}\]
\[ = \frac{22 + 42 - 12}{77} = \frac{52}{77}\]
\[ \therefore P\left( A \cup B \right) = \frac{52}{77}\]
\[\left( ii \right) P\left( \text{ none of the event occurs } \right)\]
\[ = 1 - P\left( A \cup B \right)\]
\[ = 1 - \frac{52}{77}\]
\[ = \frac{25}{77}\]