1. The problem statement, all variables and given/known data

2. Relevant equations

Σ(n*2/5*(3/5)^(n-1)=5/2

3. The attempt at a solution

First I found the number of tosses needed to get heads, but I don't understand how to interpret this in the E[X] formula.
I know that my
p(x)=.40
what is my x ? "tails for the first time" ? how to translate it in a numerical representation ?

Let's define X = number of tosses needed until the first head appears.

When x=1, it means that you get a head in your very first toss. So, p(x=1) = 0.4
When x =2, it means that you get a tail in your first toss and a head in your second toss, so p(x=2) = (0.6)(0.4).
When x = 3, ..., p(x=3) = (0.6)^2 (0.4)
...until...
When x = n, ..., p (x=n) = (0.6)^(n-1) x (0.4)

So, your E(X) = summation of x times p(x) [values of x is from 1 to infinity]
= 1 x p(x=1) + 2 x p(x=2) + 3 x p(x=3) + ... + n x p(x=n)
=1x0.4 + 2 x (0.6)(0.4) + 3 x (0.6)^2 x (0.4) + ... + n x (0.6)^(n-1) x (0.4)

Factor out 0.4.

Now, either the E(X) is a value or it might not be convergent. If it's the latter, than E(X) does not exist.

Thank you for your help, it did help me understand.
I just forgot to mention one thing. We should use calculus in our answers.
I don't think we are supposed to use geometric distribution.

However, geometric series might work here.
I just don't know what to do with the 'n'.
the sum of a geometric series is
S=a/(1-r)
the 'r' is 3/5
but I can't figure out the 'a' since it's a constant number.

The sumation is the derivative of the geometric power series [itex]\sum x^n[/itex] evaluated at 3/5. Since that series sums to 1/(1-x) then our series is 1/(1-x)^{2}.

Now that I am looking at your answer again... I don't think I understand how you are getting 1/(1-x)^2.
I understand that 1/(1-x) would be perfect here if we didn't have the 'n'
....
EDIT:

Now, that I am looking at it again, I noticed that it's the derivative of 1/(1-x).