Two dice are thrown together. The probability of getting a doublet is

(2) $\frac{1}{6}$

Explanation: 
When two dice are thrown simultaneously, all possible outcomes are:
(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)
Number of all possible outcomes  = 36

 Let E be the event of getting a doublet.
 Then the favourable outcomes are:
  (1,1), (2,2) , (3,3) (4,4), (5,5), (6,6)
    Number of favourable outcomes = 6

$\therefore P($ getting a doublet $)=P(E)=\frac{6}{36}=\frac{1}{6}$