Determining a distribution 
For part (i), I have:
1) $X$ is binomial with parameters $n$ and $p = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ i.e. $X$ is $\operatorname{Bin}(n,\frac{5}{6})$?
2) $Y$ is geometric with parameter $\frac{1}{3}$?
I think I can use the $P(Y \geq 3) = 1 - P(Y = 2) - P(Y=1) - P(Y=0)$ for the final part in that case?
Any help please
 A: Yes that is so, except that $Y$ has the support of $\{1,2,\ldots\}$. It is the count of trials until the first strictly positive result, so the count begins at $1$.   Make sure you are using the correct Geometric Distribution (there are two).   But the success rate is $1/3$.
You can also use that for such a Geometric random variable, $\Bbb P(Y\geq k)$ is the probability of $k-1$ consecutive 'failures' (ie: that the first 'success' occurs later than this).
A: Your thinking is correct, but you need to specify the distribution for $Y$ explicitly.  In particular, $\Pr[Y = 0] = 0$, because $Y$ counts the number of entries up to and including the first strictly positive entry, thus it is not possible for $Y = 0$.  The smallest value is $Y = 1$, attained with probability $\Pr[Y = 1] = 1/3$, occurring when the first draw from $\Omega$ is $1$.  With this in mind, what is $\Pr[Y = 2]$?
A: The event $Y\ge3$ is the same as failure to get a positive number on the first two trials. On each trial the probability of such failure is $2/3$, so the probability you seek is $(2/3)^2.$
