A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

On throwing a dice twice.

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)}.

No. of possible outcomes = 36

(i) Favourable outcomes for 5 to not come up either time are {(1,1) (1,2) (1,3) (1,4) (1,6) (2,1) (2,2) (2,3) (2,4) (2,6) (3,1) (3,2) (3,3) (3,4) (3,6) (4,1) (4,2) (4,3) (4,4) (4,6) (6,1) (6,2) (6,3) (6,4) (6,6)}

No. of favourable outcomes = 25

P(that 5 will not come up either time)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{25}{36}$.

Hence, the probability that 5 will not come up either time is $\dfrac{25}{36}$.

(ii) Favourable outcomes for 5 to come up at least once are {(1,5) (2,5) (3,5) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,5)}

No. of favourable outcomes = 11

P(that 5 will come up at least once)

= $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{11}{36}$.

Hence, the probability that 5 will come up at least once is $\dfrac{11}{36}$.