# 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 = $\frac{4}{4+3}=\frac{4}{7}$

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 )**

**
**