Answer:
When two number cubes are rolled,
Sample space= Total Possible outcome {11,22,33,44,55,66,12,21,13,31,14,41,15,51,16,61,23,32,24,42,25,52,26,62,34,43,35,53,36,63,45,54,46,64,56,65}=36
Probability of an event = \frac{\text{Total favorable outcome}}{\text{Total possible outcome}}
Sum of 4 is obtained when ={13,31,22}=3
Probability of getting 4 when 2 number cubes are rolled= \frac{3}{36}=\frac{1}{12}
Sum of the numbers is greater than 9= 10, 11,12={55,64,46,56,65,66}=6
Probability of getting(Sum of the numbers is greater than 9)=\frac{6}{36}=\frac{1}{6}
The sample space is the set of the 36 possible couples:
[tex] \Omega = \{(x,y): 1\leq x\leq 6,\ 1\leq y\leq 6\} [/tex]
So, we have
[tex](1,1)\ (1,2)\ (1,3)\ (1,4)\ (1,5)\ (1,6)[/tex]
[tex](2,1)\ (2,2)\ (2,3)\ (2,4)\ (2,5)\ (2,6)[/tex]
[tex](3,1)\ (3,2)\ (3,3)\ (3,4)\ (3,5)\ (3,6)[/tex]
[tex](4,1)\ (4,2)\ (4,3)\ (4,4)\ (4,5)\ (4,6)[/tex]
[tex](5,1)\ (5,2)\ (5,3)\ (5,4)\ (5,5)\ (5,6)[/tex]
[tex](6,1)\ (6,2)\ (6,3)\ (6,4)\ (6,5)\ (6,6)[/tex]
As for the probability of rolling a sum greater than 9, just count how many cases satisfy the request, and divide the number of cases by the cardinality of the sample space: the good rolls are
[tex] (3,6)\ (4,5)\ (4,6)\ (5,4)\ (5,5)\ (5,6)\ (6,3)\ (6,4)\ (6,5)\ (6,6) [/tex]
So, 10 out of 36 rolls are good, leading to a probability of
[tex] \dfrac{10}{36} = \dfrac{5}{18} [/tex]
