A dice is thrown twice. What is the probability that
1) 3 will not come up either time? 2) 6 will come up at least once?
A. \[\begin{gathered}
  (i)\dfrac{{12}}{{25}} \\
  (ii)\dfrac{{16}}{{25}} \\
\end{gathered} \]
B.\[\begin{gathered}
  (i)\dfrac{{18}}{{25}} \\
  (ii)\dfrac{9}{{25}} \\
\end{gathered} \]
C.\[\begin{gathered}
  (i)\dfrac{{25}}{{36}} \\
  (ii)\dfrac{{11}}{{36}} \\
\end{gathered} \]
D.\[\begin{gathered}
  (i)\dfrac{{23}}{{36}} \\
  (ii)\dfrac{{17}}{{36}} \\
\end{gathered} \]

Hint: Throwing a dice can have 6 outcomes. Since it is thrown twice, the final outcome would be \[{6^2}\] outcomes in total. First analyze how many times 3 shows up in all the possible outcomes. Similarly, analyze the number of times 6 shows up. Then find the probability accordingly.

Complete step-by-step answer:
When a dice is thrown twice, there are possible 36 outcomes.
3 will not come up either time, when a pair of outcomes are anything except
(1,3), (3,1), (2,3), (3,2), (4,3), (3,4), (5,3), (3,5), (6,3), (3,6), (3,3).
3 comes up 11 times.
∴P (3 will not come up either time) = no. of times 3 doesn’t show up divided by total no. of outcomes.
P (3 will not come up either time) = \[\dfrac{{36 - 11}}{{36}}\] = \[\dfrac{{25}}{{36}}\]
 7 will come up at least once, when pair of outcomes are
(1,6), (6,1), (6,2), (2,6), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5), (6,6)
6 came up 11 times.
P (6 will come up at least once) = No. of times 6 shows up at least once divided by total no. of out comes.
∴ P (6 will come up at least once) \[ = \dfrac{{11}}{{36}}\]
∴ The correct option is ‘C’.

Note: Throwing one dice twice or two dices simultaneously are treated as the same experiment. It would make the questions easy if one can write the sample space-
(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).