Option C:
![\frac{7}{36}](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B36%7D)
Solution:
Sample space for two dice
![=\left\{\begin{array}{l}{(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),(6,6)} \\{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}\end{array}\right.](https://tex.z-dn.net/?f=%3D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7D%7B%281%2C1%29%2C%281%2C2%29%2C%281%2C3%29%2C%281%2C4%29%2C%281%2C5%29%2C%281%2C6%29%7D%20%5C%5C%7B%282%2C1%29%2C%282%2C2%29%2C%282%2C3%29%2C%282%2C4%29%2C%282%2C5%29%2C%282%2C6%29%7D%20%5C%5C%7B%283%2C1%29%2C%283%2C2%29%2C%283%2C3%29%2C%283%2C4%29%2C%283%2C5%29%2C%283%2C6%29%7D%20%5C%5C%7B%284%2C1%29%2C%284%2C2%29%2C%284%2C3%29%2C%284%2C4%29%2C%284%2C5%29%2C%284%2C6%29%7D%20%5C%5C%7B%285%2C1%29%2C%285%2C2%29%2C%285%2C3%29%2C%285%2C4%29%2C%285%2C5%29%2C%286%2C6%29%7D%20%5C%5C%7B%286%2C1%29%2C%286%2C2%29%2C%286%2C3%29%2C%286%2C4%29%2C%286%2C5%29%2C%286%2C6%29%7D%5Cend%7Barray%7D%5Cright.)
Number of sample space = 36
Pair getting sum of 12 = (6, 6)
Number of pair getting sum of 12 = 1
Pair getting doublet = (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Number of pair getting doublet = 6
Total number of pair getting sum of 12 or doublet = 1 + 6 = 7
Probability of getting sum of 12 or doublet
![= \frac{\text { Total number of getting sum of } 12 \text { or doublet }}{\text { Number of sample space }}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Ctext%20%7B%20Total%20number%20of%20getting%20sum%20of%20%7D%2012%20%5Ctext%20%7B%20or%20doublet%20%7D%7D%7B%5Ctext%20%7B%20Number%20of%20sample%20space%20%7D%7D)
![=\frac{7}{36}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B7%7D%7B36%7D)
Hence,
is the probability of getting a sum of 12 or doublet when rolling a pair of dice.