Let S be the universal set
Set S :{1,2,3,4,5,6}
Set A :{your choice} $A \in\ S$
Set B :{your choice} $B \in\ S$
Exhaustive Events: You can choose any way you want to define A and B such that $A\cup\ B$ = S
Example :
|A ={1} B={2,3,4,5,6}|
|A ={1,3} B={2,4,5,6}|
|A ={1,2,5,6} B={3,4}|
|A ={1,2,3,6} B={2,3,4,5}|
|A ={1,2,3} B={2,3,4,5,6}|
|A ={1,2,3,4,5} B={1,2,3,4,5,6}| .....
Mutually Exclusive : You can choose any way you want to define A and B such that $A \cap\ B$ = $\phi\ $
i.e. There should be no common element between A and B
Example :
|A ={1} B={2,3,4,5,6}|
|A ={1,3} B={2,4,5,6}|
|A ={1,2,5,6} B={3,4}|
|A ={1,2,3} B={4,5}|
|A ={1} B={2,3}|
|A ={1,6} B={2,4,5}| .....
Equally likely: It means the probability of occurring of events A and B is same
Here probability is determined by the size of set A and B.
For equally likely both sets have the same size
Examples for equally likely
|A ={1,2,3} B={2,3,4}|
|A ={1,3} B={4,5}|
|A ={1,2} B={1,2}|
|A ={1,3,5} B={2,4,6}|
|A ={1} B={6}|
|A ={1,4,5,6} B={2,3,4,5}| .....
Let's first count how many ways are there for there to be at least $2$ of the same face, and then how many ways for $3$ of the same face.
For the first, we use complementary counting, as we can count the number of cases when all three dice show different values. This can happen in $_6P_3=6\cdot5\cdot4=120$ ways. So, the final result is $6^3-120=96$.
For the second, there are clearly only $6$ ways this can happen.
So, the final number of possibilities is $96-6=90$. We divide by the total number of possibilities, which is $216$, to get $$\frac5{12}$$
Hence, your answer is correct. I've just provided an alternate approach.
Best Answer
The probability for an odd number on the first die and a multiple of 3 on the second is indeed $\frac16$. However, multiplying by two does not give the correct answer because it counts the case where both dice show odd multiples of 3 – a "hard six" of two threes, in craps jargon – twice.
The correct answer is $2×\frac16-\frac1{36}=\frac{11}{36}$.