Imagine we have a dice that is loaded that even number are twice likely than odd number. Assume that all even numbers are equally likely, as are all odd numbers.
1) What is the probability of throwing an even number?
2) What is the probability that the thrown number is at most 4?
3) With two dice of this kind what is the probability that the combined          number of eyes shown is a most 5?
I found this answer, what do you think:
1) We know that in normal dice the probability of getting an even number is 50%, but in this case the probability will be 75%, but how can I explain it better?
2) We know that in a normal dice, the probability would have been 4/6, but we have a cheted dice, so the probability of the number 2 and 4 will be twice than a normal dice, so we will have 6/6, but this make no sense.
I am stuck with this questions. Please Help
3) I have no idea. Hint?
 A: I think you need to work through it again.  If getting an even number is twice as likely as getting an odd number, then it's as if you had a 9-sided dice with two 2's, two 4's and two 6's.  
A: *

*What is the probability of throwing an even number? 


The answer is 2/3 (not 3/4). It needs to be double that of the odd numbers, and the total must be 1.


*What is the probability that the thrown number is at most 4?  


Assuming a 6 sided dice that has 1-6 on its faces. Each odd number is 1/9, and each even is 2/9. Rolling $\leq 4$ has probability of 6/9 = 2/3.


*With two dice of this kind what is the probability that the combined number of eyes shown is a most 5?


For this, you need to look at the various combinations. The answer is 19/81. I'll leave it as an exercise, but just write out all the different combinations of 2 dice that add up to less than or equal to 5, and add up their probabilities. The combined probability for each combination is the product of the two probabilities.
A: The respective probabilities of faces 1 through 6 are $p=(1,2,1,2,1,2)/9.$
(1) $P(Even) = P\{2,4,6\} = 6/9 = 2/3.$
(2) $P\{1,2,3,4\} = (1 + 2 + 1 + 2)/9 = 6/9 = 2/3.$
(3) What do you mean 'eyes'? 'Snake-eyes' is a double-1, but that does not make sense the way I read your question. Anyway make a $6 \times  6$ array of the 36 possible pairs of
numbers, and write the appropriate probability in each cell (1/81, 2/81, or 4/81). Mark the relevant cells and add the marked probabilities. (If you mean 'total $\le 5$,' I agree with @DrXorile.)
Note: In practice, because of the way the numbers 1 through 6 are positioned
on a cubical die, it would be almost impossible to weight a die in the way you suggest. But on a crooked Internet gambling site, any assignment of probabilities is possible.
Here are results of a simulation of many rolls of such biased dice:
d1 = sample(1:6, 10^6, rep=T, prob=c(1,2,1,2,1,2))
mean(d1 <= 4) # aprx answ (1)
## 0.666425
mean(d1==2 | d1==4 | d1==6)
## 0.666845   # aprx answ (2)
d2 = sample(1:6, 10^6, rep=T, prob=c(1,2,1,2,1,2))
s = d1+d2     # totals on two dice
mean(s <= 5)
## 0.23438    # aprx answ (3)
19/81
## 0.2345679  # exact answ (3)

Below is a histogram of the simulated sums when two such biased dice are rolled.
Horizontal grid lines show multiples of $1/81.$

