Explanation

(b)
We have, $P(S) = P\left\{ 5\ ,6 \right\} = \frac{2}{6} = \frac{1}{3}$
Let us denote the occurrence of a number greater than 4 in a single throw of the die and $F$ denote its failure.
$\Rightarrow P(F) = \frac{2}{3}$
$P$ (an even number of tosses is needed)
$= P(FS\ or\ FFFS\ or\ FFFFFS\ or\ldots)$
$= P(F)P(S) + P(F)^{3}P(S) + P(F)^{5}P(S) + ...$
$= \frac{P(F)P(S)}{1 - P(F)^{2}} = \frac{\frac{2}{9}}{1 - \frac{4}{9}} = \frac{2}{5}$