Explanation

(b)
$p = Probability\ of\ success\ (s) = \frac{2}{6} = \frac{1}{3}$
$q = Probability\ of\ failure\ (f) = 1 - \frac{1}{3} = \frac{2}{3}$
Probability that success occurs in even number of tosses
$= P(fs) + P(f\ f\ fs) + P(f\ f\ f\ f\ fs) + \ ...$
$= qp + q^{3}p + q^{5}p + ... = \frac{qp}{1 - q^{2}}$
$= \frac{\frac{2}{3} \times \frac{1}{3}}{1 - \left( \frac{2}{3} \right)^{5}} = \frac{\frac{2}{9}}{1 - \frac{4}{9}} = \frac{2}{9} \times \frac{9}{5} = \frac{2}{5}$