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

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$\frac{2}{5}$

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(B) The probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss is $p = \frac{2}{6} = \frac{1}{3}$.The probability of failure (getting $1, 2, 3,$ or $4$) is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.We need the probability that the success occurs on an even number of tosses,which corresponds to the sequence of outcomes: (Failure,Success),(Failure,Failure,Failure,Success),(Failure,Failure,Failure,Failure,Failure,Success),and so on.The required probability is $P = pq + q^3p + q^5p + \dots$This is an infinite geometric series with the first term $a = pq$ and common ratio $r = q^2$.The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.Substituting the values: $P = \frac{pq}{1 - q^2} = \frac{(\frac{1}{3})(\frac{2}{3})}{1 - (\frac{2}{3})^2} = \frac{\frac{2}{9}}{1 - \frac{4}{9}} = \frac{\frac{2}{9}}{\frac{5}{9}} = \frac{2}{5}$.

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$4k$	$3k$	$k$

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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