Three ordinary and fair dice are rolled simultaneously. The probability of the sum of outcomes being atleast equal to 8, is equal to

$\frac{181}{216}$

Let the outcome on dice be n₁, n₂ and n₃. we first try to find the number of ways in which sum of outcomes is at the most equal to seven.

For this, $n_1+n_2+n_3 \leq 7$, where $n_i \in[1,6]$

$\Rightarrow\left(n_1-1\right)+\left(n_2-1\right)+\left(n_3-1\right)+y_4=4$

i.e.,  $y_1+y_2+y_3+y_4=4$

where $y_1, y_2, y_3, y_4$ are non-negative integers. This can happen in ${ }^{4+4-1} C_4$ i.e., ${ }^7 C_4$ ways.

Thus, required probability

$=1-\frac{{ }^7 C_4}{6^3}=\frac{181}{216}$