A) \[\frac{2}{9}\]
B) \[\frac{3}{7}\]
C) \[\frac{4}{9}\]
D) \[\frac{5}{9}\]
Correct Answer: A
Solution :
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 \[=\frac{8}{36}=\frac{2}{9}\]
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