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(B) Let $p$ be the probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss.$p = \frac{2}{6} = \frac{1}{3}$.Let $q$ be the

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$\frac{2}{5}$

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(B) Let $p$ be the probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss.$p = \frac{2}{6} = \frac{1}{3}$.Let $q$ be the probability of failure (i.e.,getting $1, 2, 3,$ or $4$).$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.We need the probability that the success occurs on an even number of tosses (i.e.,$2^{nd}, 4^{th}, 6^{th}, \dots$ toss).The probability is given by the sum of the infinite geometric series:$P = P(FS) + P(FFFS) + P(FFFFFS) + \dots$$P = qp + q^3p + q^5p + \dots$This is a geometric progression with first term $a = qp$ and common ratio $r = q^2$.The sum of an infinite geometric series is $S = \frac{a}{1-r}$.$P = \frac{qp}{1 - q^2} = \frac{(\frac{2}{3})(\frac{1}{3})}{1 - (\frac{2}{3})^2} = \frac{\frac{2}{9}}{1 - \frac{4}{9}} = \frac{\frac{2}{9}}{\frac{5}{9}} = \frac{2}{5}$.

An unbiased die is tossed until a number greater than $4$ appears. The probability that an even number of tosses is needed is

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