A fair die is tossed repeatedly until a six is obtained. Let X...

P\left(\text{getting}6\right) = \dfrac{1}{6}, P\left(\text{not getting}6\right) = \dfrac{5}{6}$</p><p>$⇒ a = P ( X = 3 )$</p><p>$⇒ a = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6}$</p><p>$⇒ a = \dfrac{25}{216} . . . \left(i\right)$</p><p>$⇒ b = P ( X \ge 3 )$</p><p>$⇒ b = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} + {\left(\dfrac{5}{6}\right)}^{3} \cdot \dfrac{1}{6} + {\left(\dfrac{5}{6}\right)}^{4} \cdot \dfrac{1}{6} + . . .$</p><p>We know that,${S}_{\infty} = a + ar + {ar}^{2} + . . . \infty = \dfrac{a}{1 - r}$</p><p>$⇒ b = \dfrac{\dfrac{25}{216}}{1 - \dfrac{5}{6}}$</p><p>$⇒ b = \dfrac{25}{216} \times \dfrac{6}{1}$</p><p>$⇒ b = \dfrac{25}{36} . . . \left(ii\right)$</p><p>Now,$P ( X \ge 6 ) = {\left(\dfrac{5}{6}\right)}^{5} \cdot \dfrac{1}{6} + {\left(\dfrac{5}{6}\right)}^{6} \cdot \dfrac{1}{6} + . . .$</p><p>$⇒ P ( X \ge 6 ) = \dfrac{{\left(\dfrac{5}{6}\right)}^{5} \cdot \dfrac{1}{6}}{1 - \dfrac{5}{6}}$</p><p>$⇒ P ( X \ge 6 ) = {\left(\dfrac{5}{6}\right)}^{5}$</p><p>$⇒ c = P\left(\dfrac{X \ge 6}{X > 3}\right) = \dfrac{P\left(\left(X \ge 6\right) \cap \left(X > 3\right)\right)}{P\left(X > 3\right)} = \dfrac{P\left(X \ge 6\right)}{P\left(X > 3\right)} = \dfrac{{\left(\dfrac{5}{6}\right)}^{5}}{{\left(\dfrac{5}{6}\right)}^{3}}$</p><p>$⇒ c = \dfrac{25}{36} . . . \left(iii\right)$</p><p>Using \left(i\right), \left(ii\right)\text{and}\left(iii\right)$,

$⇒ \dfrac{b + c}{a} = \dfrac{\dfrac{25}{36} + \dfrac{25}{36}}{\dfrac{25}{216}}$

$⇒ \dfrac{b + c}{a} = \dfrac{\dfrac{50}{36}}{\dfrac{25}{216}}$

$⇒ \dfrac{b + c}{a} = 12$