Average number of throws of 2 dice in order for the sum to be a prime number If I throw 2 dice, what's the average number of throws that I need to do in order for the sum of the 2 dice to be a prime number?
My attempt:
The probability that the sum will be a prime number is $\frac{4}{9}$.  And, by applying the formula of mathematical expectation($\frac{1}{p}$) I get $\frac{1}{\frac{4}{9}}$ then $\frac{9}{4} > 1$. Which is not correct.
 A: As A. Webb notes above, check your assumption that the sum being prime is 4/9ths. Two independent dice of six faces give you thirty six different results, many of which turn into the same sums, and since that's a very countable number actually making a chart can be really helpful. 
Now, I'm not sure what you are saying with the formula of mathematical expectation (between the original question and your edit) but I believe it is  that you have some number of independent throws which each have some chance 'p' of being a prime. I think you are trying to use the reciprocal of that (1/p) to give you the chance of it NOT being prime.
That's not how you get there. You have the chance of the sum being a prime 'p,' what you want is its complement: one minus p. What's that mean? One in probability is a 100% chance, so one minus "the chance of a prime" is "the chance of it NOT being prime" that I think you're looking for as your intermediate step.
If that isn't your intention, feel free to comment.
A: Prime numbers between $2$ and $12$ are $2,3,5,7,11$
which can be obtained in $15$ ways as $\boxed{1-1}\boxed{ 1-2/2-1}\boxed{1-4/4-1,\; 2-3/3-2}\boxed{1-6/6-1,2-5/5-2,3-4/4-3}\boxed{ 5-6/6-5} $ 
Thus P(prime number) = $\dfrac{15}{36} = \dfrac5{12}$
and $E[X] = 1/p = \dfrac{12}5 = 2.4$
