Probabilites of throwing two six-sided, loaded dice I am trying to calculate some probabilities of throwing two loaded six-sided dice.
The probabilities of odd numbers are three times that of even numbers, so:
p(2) = p(4) = p(6) = x => 3x = p(1) = p(3) = p(5)

the probability of sample space should be 1
   p(1)+p(2)+p(3)+p(4)+p(5)+p(6) = 1
=> 3x  + x  + 3x + x  + 3x + x   = 1
=> x = 1/2
=> p(even number) = 1/12 and p(odd number) = 1/4

now, I am asked to find the probabilities of 
1) the sum of the two dice to be 6
let E be the event where the sum of the two dice roll adds up to 6
E = {,,,,}
p(E) = p() + p() + p() + p() + p()
p(E) = p()*p() + p()*p() + p()*p() + p()*p() + p()*p() 
p(E) = 1/4*1/4 + 1/12*1/12 + 1/4*1/4 + 1/12*1/12 + 1/4*1/4 = 29/144

2) dice doubles
let D be the event where the two dice roll ends up in doubles
D = {,,, ,,}
p(D) = p() + p() + p() + p() + p() + p()
p(D) = p()*p() + p()*p() + p()*p() + p()*p() + p()*p() + p()*p() 
p(D) = 1/4*1/4 + 1/12*1/12 + 1/4*1/4 + 1/12*1/12 + 1/4*1/4 + 1/12*1/12 = 5/24

3) at least one dice is 2
let T be the event where at least a 2 comes up, trought the same method:
p(T) = 23/144
4) this is the question that I can't answer, supposing all above is right, the probability of the sum being 6 or doubles and to be at least a 2, I am interpreting this as 
   p[(E ∪ D) ∩ T], with De Morgan 
=>  p(E ∩ T) ∪ p(D ∩ T)    
=   p(E ∩ T) + p(D ∩ T) 
=   p(E)*p(T) + p(D)*p(T)
=   29/144*23/144 + 5/24*23/144
=   1357/20736 
≈   0.0654

now, I calculated p[(E ∪ D) ∩ T] another way, first p(E ∪ D) = 25/72, this way it would just be 25/72 * 23/144 which equals 575/10368 ≈ 0.055
clearly 575/10368 ≠ 1357/20736
don't understand what I am doing wrong, would appreciate any help
 A: The answer below is with $T$ defined as the event where at least one dice is equal or greater than $2$.
You cannot proceed as stated because $E$ and $D$ are not disjoint: The outcome $(3,3)$ sums to $6$ and is a double. Thus $P(D\cup E) \neq P(D) + P(E)$, and similarly, this might not even be the case when you intersect with $T$.
But with $D^*$ being the doubles excluding $(3,3)$, you have that $D^*$ and $E$ are disjoint while also $D\cup E = D^*\cup E$. Thus
$P((D\cup E) \cap T) = P((D^*\cup E) \cap T) = P((D^* \cap T)\cup D (E \cap T))$,
using De Morgan's laws. As $D^*$ and $E$ are disjoint, so are $D^* \cap T$ and $D^* \cap E$, hence
$P((D\cup E) \cap T) = P(D^* \cap T) + P(E \cap T)$.
Here $(D^* \cap T) = \{(2,2),(4,4),(5,5),(6,6)\}$. I think you can proceed from here on, following your line of thoughts from the earlier problems. 
A final remark: $P(A \cap B) = P(A) \cdot P(B)$ only holds when $A$ and $B$ are independent events. It is not a general property, that one can use at one's discretion.
If $T$ is defined as the set where at least one dice is equal to $2$, then $(D^* \cap T)=(2,2)$ instead.
A: First, for simplicity, I suggest you work the problem for
two fair dice. Make the obvious $6 \times 6$ table of the 36
possible outcomes, and count favorable outcomes for your
events $E,$ $D,$ and $T,$ which have probabilities 5/36, 1/6,
and 11/36, respectively. Then, you can see that the event $(E \cup D)\cap T)$
has three favorable outcomes and hence probability 3/36.
Here is a simulation of 1 million iterations of your experiment, which
should approximate probabilities to about two-place accuracy.
[Notes on R code: The m-vectors E, D, and T are logical vectors, consisting of TRUEs
and FALSEs, and the mean of a logical vector is the proportion of its
TRUEs; also | means or ($\cup$) and & means and ($\cap$).
Also, the name T is reserved for other purposes in R, so I used T2.]
 m = 10^6
 d1 = sample(1:6, m, rep=TRUE)  # m rolls of the 1st die
 d2 = sample(1:6, m, rep=TRUE)  # 2nd
 E = (d1 + d2 == 6)
 mean(E)
 ## 0.138751
 5/36
 ## 0.1388889
 D = (d1==d2)
 mean(D)
 ## 0.16601
 1/6
 ## 0.1666667
 T2 = (d1==2 | d2==2)
 mean(T2)
 ## 0.305209
 11/36
 ## 0.3055556
 mean((E|D)&T2)
 ## 0.083279
 3/36
 ## 0.08333333

Now here is a modification of the simulation for your unfair dice.
You can use the results to check (approximately) your answers for $E,$ $D,$ and $T$.
Also, for the final result, when you get it.
 m = 10^6;  pr = rep(c(1/4, 1/12), 3)
 d1 = sample(1:6, 10^6, rep=TRUE, prob=pr)
 d2 = sample(1:6, 10^6, rep=TRUE, prob=pr)
 E = (d1 + d2 == 6)
 mean(E)
 ## 0.201538
 D = (d1==d2)
 mean(D)
 ## 0.208269
 T2 = (d1==2 | d2==2)
 mean(T2)
 ## 0.159524
 mean((E|D)&T2)
 ## 0.020934
 3/144
 ## 0.02083333

