In a throw of a dice,the probability of getti | vedclass

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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 $q = 1 - \frac{1}{6} = \frac{5}{6}$.The event of getting a $1$ in an even number of throws means getting it on the $2^{nd}, 4^{th}, 6^{th}, \dots$ throw.This is represented by the sum of the probabilities: $q \cdot p + q^3 \cdot p + q^5 \cdot p + \dots$This is an infinite geometric series with the first term $a = qp = \frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}$ and common ratio $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.The sum $S$ of an infinite geometric series is given by $S = \frac{a}{1 - r}$.$S = \frac{5/36}{1 - 25/36} = \frac{5/36}{11/36} = \frac{5}{11}$.

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

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