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(B) Let $p$ be the probability of getting a $1$ in a single throw,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a $1$,so $q = 1 - \fr

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$\frac{5}{11}$

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(B) Let $p$ be the probability of getting a $1$ in a single throw,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a $1$,so $q = 1 - \frac{1}{6} = \frac{5}{6}$.We want to find the probability of getting a $1$ for the first time on an even throw (i.e.,$2^{nd}, 4^{th}, 6^{th}, \dots$ throw).The probability is given by the sum of the infinite geometric series:$P = qp + q^3p + q^5p + \dots$This is a geometric series with first term $a = qp = \frac{5}{6} 	imes \frac{1}{6} = \frac{5}{36}$ and common ratio $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.$P = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{36 - 25}{36}} = \frac{5}{36} 	imes \frac{36}{11} = \frac{5}{11}$.

In a throw of a dice,the probability of getting a $1$ in an even number of throws is:

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