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A and B throw a pair of dice alternately. A wins the game, if he gets a total of 6 and B wins, if she gets a total of 7 . If A starts the game, then find the probability of winning the game by A in third throw of the pair of dice.
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Let $\mathrm{A}$ and $\mathrm{B}$ be the event of getting a total of 6 and 7 respectively then
A total of $6=\{(2,4),(1,5),(5,1),(4,2),(3,3)\}$
A total of $7=\{(2,5),(1,6),(6,1),(5,2),(3,4)$, $(4,3)\}$
$$
\mathrm{P}(\mathrm{A})=\frac{5}{36} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{1}{6}
$$
$\therefore$ Required probability
$$
=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\mathrm{A})=\frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}=\frac{775}{7776}
$$
A total of $6=\{(2,4),(1,5),(5,1),(4,2),(3,3)\}$
A total of $7=\{(2,5),(1,6),(6,1),(5,2),(3,4)$, $(4,3)\}$
$$
\mathrm{P}(\mathrm{A})=\frac{5}{36} \Rightarrow \mathrm{P}(\mathrm{B})=\frac{1}{6}
$$
$\therefore$ Required probability
$$
=\mathrm{P}(\overline{\mathrm{A}}) \cdot \mathrm{P}(\overline{\mathrm{B}}) \cdot \mathrm{P}(\mathrm{A})=\frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}=\frac{775}{7776}
$$
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