Two dices are rolled. If both dices have six faces numbered 1, 2,...

n\left(S\right) = 36$</p><p>The possible ordered pairs are \left(1, 1\right), \left(1, 2\right), \left(1, 3\right), \left(1, 5\right), \left(1, 7\right), \left(2, 1\right), \left(2, 2\right), \left(2, 3\right), \left(2, 5\right), \left(3, 1\right), \left(3, 2\right), \left(3, 3\right), \left(3, 5\right), \left(5, 1\right)\left(5, 2\right), \left(5, 3\right), \left(7, 1\right)$

$n\left(E\right) = 5 + 4 + 4 + 3 + 1 = 17$

So, $P\left(E\right) = \dfrac{17}{36}$