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

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