• # question_answer Two dice are thrown simultaneously. The probability of getting a doublet or a total of 4 is ___. A)  $\frac{2}{9}$                       B)  $\frac{3}{7}$           C)  $\frac{4}{9}$                       D)  $\frac{5}{9}$

Total number of outcomes, when two dice are thrown $=6\times 6=36$ Total number of doublets present = (1, 1), (2, 2), (3, 3), (4,4). (5, 5), (6, 6) For a total of 4, pairs can be (1, 3), (3, 1), (2, 2). Thus, total number of favourable outcomes = 8 [since (2, 2) is present in both the cases], $\therefore$ Required probability $=\frac{8}{36}=\frac{2}{9}$