Expected number of virus cells I've found this question in a past programming assignment from a course I'm currently reading.
Its statement looks like this :

A recent lab accident resulted in the creation of an extremely dangerous virus that replicates so rapidly it's hard to predict exactly how many cells it will contain after a given period of time. However, a lab technician made the following observations about its growth per millisecond:
$\bullet$ The probability of the number of virus cells growing by a factor of $a$ is $0.5$
$\bullet$ The probability of the number of virus cells growing by a factor of $b$ is $0.5$
Given a, b, and knowing that initially there is only a single cell of virus, calculate the expected number of virus cells after $t$ milliseconds. As this number can be very large, print your answer modulo $(10^9 + 7)$ .

As I have no prior training in probability or combinatorics, this problem doesn't make much sense to me . I've done some searching about expected values in the context of probability, but I can't see how to model the data I'm given. Perhaps there's something very obvious I'm missing, but I'm not able to see it at the moment.
How would you solve this?

Let $X$ be the number of times the "virus cell" population grew by factor $a$ in the $t$ milliseconds. 
Thus the population of 1, after $X$ growths of factor $a$ and $t-X$ growths of factor $b$ will be: $$\begin{align}N_t =&~ a^X b^{t-X} \\ = &~ b^t (a/b)^{X} \end{align}$$
Now, as $X$ is is the count of 'successes' in a series of $t$ independent Bernoulli trials with success rate $0.5$, it will have a Binomial Distribution.   $X\sim\mathcal {Bin}(t, 0.5)$
Now can you find this?  $~\mathsf E(N_t)$

Hint: A useful research topic is "moment generating functions".

 If $X\sim \mathcal{Bin}(n,p)$ then $\mathsf E(\mathsf e^{sX}) = (1-p+p\mathsf e^{s})^n$.   This is the moment generating function for a Binomial Distribution.

