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 ________.

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

So,
Probability of winning A =
And
Probability of not winning A =
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 -
So,
Odds in favour of A winning at least once  =  316 :  343                                                                                       ( Ans )

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