A die is thrown twice. What is the probability that

(i) 3 will not come up either time?

(ii) 6 will come up at least once?

The probability of an event E occurring is the ratio of the number of cases in its favour to the total number of equally likely cases.

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{\text{ no. of cases favourable to event }E}{\text{ Total no. of cases }}$

When a dice is thrown twice, the possible 36 outcomes.

$\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}$

(i) The number of times that 3 would come up is 11, which can be seen as follows:$(1,3),(3,1),(3,3),(2,3),(3,2),(3,6),(6,3)(4,3),(3,4),(3,5),(5,3)$

Therefore, the probability that 3 will not come up either time $=\frac{36-11}{36}=\frac{25}{36}$

(ii) 6 will come up at least once, when the pair of outcomes are:

$(1,6),(6,1),(6,2),(2,6),(3,6),(6,3),(4,6),(6,4),(6,5),(5,6),(6,6)$

Probability that 6 will come up at least once $=\frac{11}{36}$