The odds against an event are 5 : 3 and the odds in favour of another independent event are 7 : 5. Find the probability that at least one of the two events will occur.

Solution (By Examveda Team)

Let probability of the first event taking place be A and probability of the second event taking place be B.
Then
$$\eqalign{ & P\left( A \right) = \frac{3}{{5 + 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{8} \cr & P\left( B \right) = \frac{7}{{7 + 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{7}{{12}} \cr} $$
The required event can be defined as that A takes place and B does not take place (A or B takes place and A does not take place or A takes place and B takes place.)
$$ = \left[ {P\left( A \right)\left\{ {1 - P\left( B \right)} \right\}} \right] + $$     $$\left[ {P\left( B \right)\left\{ {1 - P\left( A \right)} \right\}} \right] + $$     $$\left[ {P\left( A \right)P\left( B \right)} \right]$$
$$ = \left[ {\frac{3}{8} \times \frac{5}{{12}}} \right] + $$   $$\left[ {\frac{5}{8} \times \frac{7}{{12}}} \right] + $$   $$\left[ {\frac{3}{8} \times \frac{7}{{12}}} \right]$$
$$\eqalign{ & = \frac{{15 + 35 + 21}}{{96}} \cr & = \frac{{71}}{{96}} \cr} $$