Finding mean from die probability 
Example 4.4.5: Suppose that there is a 6-sided die that is weighted in
  such a way that each time the die is rolled, the probabilities of
  rolling any of the numbers from 1 to 5 are all equal, but the
  probability of rolling a 6 is twice the probability of roll- ing a 1.
  When you roll the die once, the 6 outcomes are not equally likely.
  What are the probabilities of the 6 outcomes?

On the basis of the above question  a question has been asked to evaluate the mean.

My problem is that, when they  evaluate the mean, they multiplied with the probability by 1, 2 ,  3  , 4 , 5 and  6 why?  In die , all are they equal probable.  
 A: The mean of a discrete random variable is defined as 
$$\mathbb{E}(X)=\sum\limits_{x\in X} xp(x)$$
In this case $X=\{1,2,3,4,5,6\}$, so that's where you are getting the multiplication. You can think of the probabilities as weighting the importance of each of these $6$ numbers.
A: As Peter stated, to calculate the mean, you multiply each outcome (the number that you roll) by the probability of getting that roll, and add those products up.
On a regular or fair die, yes, the probabilities of each outcome are the same.  But this isn't a fair die; $6$ is twice as probable as each of the other numbers.
For a fair die, $$E(X) = \frac{1}{6}(1) + \frac{1}{6}(2) + \frac{1}{6}(3) + \frac{1}{6}(4) + \frac{1}{6}(5) + \frac{1}{6}(6) = 3.5.$$
For the unfair die in your problem, $E(X)$ is as calculated.
A: The best and naivest way to understand this is to imagine a 7 sided dice with 2 faces holding the value 6. By the definition of the mean you multiply each element of the set by it's probability ... hence 1/7 to each.
