Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6 , are...

Perfect Square can be $4$ or $9$, which can be obtained by \left(1, 3\right), \left(3, 1\right), \left(2, 2\right), \left(6, 3\right)\left(3, 6\right), \left(5, 4\right), \left(4, 5\right)$.</p><p>Prime Number can be$2$or$3$or$5$or$7$or$11$, which can be obtained by </p><p>\left(1, 1\right), \left(1, 2\right), \left(2, 1\right), \left(1, 4\right), \left(4, 1\right), \left(2, 3\right), \left(3, 2\right), \left(1, 6\right), \left(6, 1\right), \left(5, 2\right), \left(2, 5\right), \left(3, 4\right), \left(4, 3\right), \left(6, 5\right), \left(5, 6\right)$

$P$(square or prime)$= \dfrac{22}{36}$

$P\left(\dfrac{\text{Perfect square is odd}}{\text{perfect square before prime}}\right) = \dfrac{\dfrac{4}{36} + \dfrac{14}{36} . \dfrac{4}{36} + \dfrac{14}{36} . \dfrac{14}{36} . \dfrac{4}{36} + . . . .}{\dfrac{7}{36} + \dfrac{14}{36} . \dfrac{7}{36} + \dfrac{14}{36} . \dfrac{14}{36} . \dfrac{7}{36} + . . . . . .}$

$= \dfrac{\left(\dfrac{\dfrac{4}{36}}{1 - \dfrac{14}{36}}\right)}{\left(\dfrac{\dfrac{7}{36}}{1 - \dfrac{14}{36}}\right)} = \dfrac{4}{7} = P$

$⇒ 14P = 8$