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If two biased dice are rolled simultaneously until a sum of the number appeared on these dice is either 7 or 11 , then the probability that 7 comes before 11 , is
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Verified Answer
The correct answer is:
$\frac{3}{4}$
Let $A$ be the events sum appeared on two unbiased dice is 7 and $B$ be the event sum appeared on two unbiased dice is either 7 or 11 .
$$
\therefore P(A)=\frac{6}{36}, P(B)=\frac{6}{36}+\frac{2}{36}=\frac{2}{9}
$$
$\therefore$ Required probability
$$
\begin{aligned}
& =P(A)+P(\bar{B} A)+P(\bar{B} \bar{B} A)+P(\overline{B B B} A)+\ldots \\
& =\frac{1}{6}+\frac{7}{9} \times \frac{1}{6}+\left(\frac{7}{9}\right)^2 \times \frac{1}{6}+\ldots \\
& =\frac{1}{6}\left(\frac{1}{1-\frac{7}{9}}\right)=\frac{1}{6} \times \frac{9}{2}=\frac{3}{4}
\end{aligned}
$$
$$
\therefore P(A)=\frac{6}{36}, P(B)=\frac{6}{36}+\frac{2}{36}=\frac{2}{9}
$$
$\therefore$ Required probability
$$
\begin{aligned}
& =P(A)+P(\bar{B} A)+P(\bar{B} \bar{B} A)+P(\overline{B B B} A)+\ldots \\
& =\frac{1}{6}+\frac{7}{9} \times \frac{1}{6}+\left(\frac{7}{9}\right)^2 \times \frac{1}{6}+\ldots \\
& =\frac{1}{6}\left(\frac{1}{1-\frac{7}{9}}\right)=\frac{1}{6} \times \frac{9}{2}=\frac{3}{4}
\end{aligned}
$$
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