In a single throw of a pair of dice, the probability of getting the sum a perfect square is
(a) $\frac{1}{18}$
(b) $\frac{7}{36}$
(c) $\frac{1}{6}$
(d) $\frac{2}{9}$
GIVEN: A pair of dice is thrown
TO FIND: Probability of getting the sum a perfect square
Let us first write the all possible events that can occur
(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),
Hence total number of events is 
Favorable events i.e. getting the sum as a perfect square are
(1,3), (2,2), (3,1), (3,6), (4,5), (5,4), (6,3)
Hence total number of favorable events is 7
We know that PROBABILITY = 
Hence probability of getting the sum a perfect square is
Hence the correct option is option 