Question

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?
** [Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]**.

Answer 1

Total number of outcomes = 6 × 6 = 36

(i) Total number of outcomes when 5 comes up on either time are (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)
Hence, total number of favorable cases = 11
P (5 will come up either time) =$\frac{11}{36}$
P (5 will not come up either time) $=1-\frac{11}{36}=\frac{25}{36}$

(ii) Total number of cases, when 5 can come at least once = 11
P (5 will come at least once) =$\frac{11}{36}$