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It has been found that if $A$ and $B$ play a game 12 times, $A$ wins 6 times, $B$ wins 4 times and they draw twice. $A$ and $B$ take part in a series of 3 games. The probability that they win alternately, is :
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The correct answer is:
$5 / 36$
$\mathrm{P}(\mathrm{A})=\frac{6}{12}=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{4}{12}=\frac{1}{3}$
Req. probability $=\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2}+\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3}=\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$
Req. probability $=\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2}+\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3}=\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$
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