Two dice are thrown simultaneously. What is the probability that :

(i) 4 will not come up either time ?

(ii) 4 will come up at least once ?

When two dice are thrown simultaneously;

No. of possible outcomes = 6 × 6 = 36.

(i) Favourable outcomes for 4 not coming on any of the dice are : (1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 5), (6, 6).

No. of favourable outcomes = 25

P(such that 4 will not come up either time)

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

Hence, the probability of not getting 4 any time = $\dfrac{25}{36}$.

(ii) Favorable outcomes for 4 coming up at least once are :

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

No. of favourable outcomes = 11

P(such that 4 will come up at least once)

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

Hence, the probability of 4 coming up at least once = $\dfrac{11}{36}$.