Answer:
[tex]\dfrac{1}{12}[/tex]
Step-by-step explanation:
All possible outcomes:
[tex]\begin{array}{cccccc}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\end{array}[/tex]
Here, the first number is the number on the 1st die and the second number is the number on the 2nd die.
There are 36 possible outcomes.
A total of 4 give outcomes:
[tex]1,3\\2,2\\3,1[/tex]
So, there are 3 favorable outcomes.
The probability of getting a total of 4 is
[tex]\dfrac{3}{36}=\dfrac{1}{12}[/tex]
