The odds in favour of A winning a game against B is 4:3. If three games to be played to decide the overall winner, the odds in favour of A winning at least once is ________.

Answer :

Given  :  The odds in favour of A winning a game against B is 4 : 3 

So,
Probability of winning A =  $\frac{4}{4 + 3} = \frac{4}{7}$ 
And
Probability of not winning A =  $1 - \frac{4}{7} = \frac{7 - 4}{7} = \frac{3}{7}$
So,
Probability of winning at least one game by A  =  1  -  Probability of winning no game

Probability of winning at least one game by A  =  1  -  ${\left(\frac{3}{7}\right)}^3$ =   1 - $\frac{27}{343} = \frac{343 - 27}{343} = \frac{\mathbf{316}}{\mathbf{343}}$
So,
Odds in favour of A winning at least once  =  316 :  343                                                                                       ( Ans )