You are correct on all accounts.

To a very good approximation, molecules can be thought as made of elements (with their respective isotope distributions) combining completely independently. You can think of it like rolling multiple die at once. This means that a simple multinomial distribution will describe this problem mathematically.

Let's start with something easy and consider the hypothetical molecule $\ce{C_5}$. Furthermore, let us consider that the only carbon isotopes with significant natural occurrence are $\ce{^{12}C}$ (98.9%) and $\ce{^{13}C}$ (1.1%). We can find all of the isotopic peaks and their relative abundances by then expanding the binomial $(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5$, where $m[^{12}C]$ and $m[^{13}C]$ denote the exact masses of the carbon-12 and carbon-13 isotopes, respectively. Expanding the binomial yields:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5 ={} & \ \ \ \ \ \binom {5} {0}(0.989\times m[^{12}C])^5 \\ & + \binom {5} {1}(0.989\times m[^{12}C])^4 \times (0.011\times m[^{13}C]) \\ & + \binom {5} {2}(0.989\times m[^{12}C])^3 \times (0.011\times m[^{13}C])^2 \\ & + \binom {5} {3}(0.989\times m[^{12}C])^2 \times (0.011\times m[^{13}C])^3 \\ & + \binom {5} {4}(0.989\times m[^{12}C]) \times (0.011\times m[^{13}C])^4 \\ & + \binom {5} {5} (0.011\times m[^{13}C])^5 \\ \end{aligned} \end{equation}$

Calculating the coefficients in each term:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times (m[^{13}C])^5 ={} & \ \ \ \ \ 0.946 \times (m[^{12}C])^5 \\ & + 0.0526\times (m[^{12}C])^4 \times m[^{13}C] \\ & + 0.00117 \times (m[^{12}C])^3 \times (m[^{13}C])^2 \\ & + 0.0000130 \times (m[^{12}C])^2 \times (m[^{13}C])^3 \\ & + 7.24×10^{-8}\times (m[^{12}C]) \times ( m[^{13}C])^4 \\ & + 1.61×10^{-10} \times m[^{13}C])^5 \\ \end{aligned} \end{equation}$

From the expansion, we see that 94.6% of all $\ce{C_5}$ molecules contain only carbon-12 (the lowest possible mass for the molecule), and almost all of the rest (5.3% out of the remaining 5.4%) is accounted for by molecules that contain a single carbon-13 atom. Only about 0.1% of $\ce{C_5}$ molecules contain two or more carbon-13 atoms.

But what happens in very large molecules? Intuitively, if there are many atoms, you would expect a higher change of there being at least one less common isotope in the mix. Let's see the first few terms for the molecule $\ce{C_100}$:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ \binom {100} {0}(0.989\times m[^{12}C])^{100} \\ & + \binom {100} {1}(0.989\times m[^{12}C])^{99} \times (0.011\times m[^{13}C]) \\ & + \binom {100} {2}(0.989\times m[^{12}C])^{98} \times (0.011\times m[^{13}C])^2 \\ & + \ ...\\ \end{aligned} \end{equation}$

Calculating the coefficients:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ 0.331 \times (m[^{12}C])^{100} \\ & + 0.368 \times (m[^{12}C])^{99} \times (m[^{13}C]) \\ & + 0.203 \times (m[^{12}C])^{98} \times (m[^{13}C])^2 \\ & +\ ...\\ \end{aligned} \end{equation}$

Well that's interesting. Now only 33.1% of the molecules contain only carbon-12 atoms, and in fact more molecules contain exactly one carbon-13 atom, at 36.8% of the total. Even molecules with two carbon-13 atoms are quite abundant, at 20.3%.

Indeed, peaks containing rarer isotopes eventually dominate. For the huge molecule $\ce{C_10000}$, the strongest mass spectrum signal would come from molecules contaning 110 carbon-13 atoms, corresponding to 3.8% of the total, while a measly $9.2\times 10^{-47}\%$ of molecules contain only carbon-12. This happens because when $n$ is large, the term $\binom {n} {k}$ grows very quickly as $k$ rises from zero, overwhelming the increase in the exponent of the rarer isotope. You can see this behaviour quite nicely in this sequence of mass spectra of molecules with increasing size.

To calculate the specific $M/M+2$ ratio for a molecule containing only $n$ carbon atoms, all you need is to get the ratio for first and third terms in the binomial:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^n ={} & \ \ \ \ \ \color{#0000ff}{ \binom {n} {0}(0.989\times m[^{12}C])^n} \\ & + \binom {n} {1}(0.989\times m[^{12}C])^{n-1} \times (0.011\times m[^{13}C]) \\ & + \color{#0000ff}{\binom {n} {2}(0.989\times m[^{12}C])^{n-2} \times (0.011\times m[^{13}C])^2} \\ & +\ ...\\ \end{aligned} \end{equation}$

The ratio is then:

$$\frac{\binom {n} {0}0.989^n}{\binom {n} {2}0.989^{n-2} \times 0.011^2}=\frac{2\times 0.989^2}{n(n-1)\times 0.011^2}$$

Technically this only holds if there are no other elements which contain multiple isotopes, though it will hold approximately if the other elements only have very rare alternate isotopes, such as hydrogen (99.98% hydrogen-1, 0.02% hydrogen-2).

As a last curiosity, all of the above extends to analysing more complicated molecules. For example, glucose ($\ce{C6H12O6}$) will have a mass spectrum exactly described by the expression:

$$(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^6 \times (0.9998\times m[^{1}H] + 0.002\times m[^{2}H])^{12} \times (0.9976\times m[^{16}O] + 0.004\times m[^{17}O] + 0.020\times m[^{18}O])^6$$

Happy expanding!

You are correct on all accounts.

To a very good approximation, molecules can be thought as made of elements (with their respective isotope distributions) combining completely independently. You can think of it like rolling multiple die at once. This means that a simple multinomial distribution will describe this problem mathematically.

Let's start with something easy and consider the hypothetical molecule $\ce{C_5}$. Furthermore, let us consider that the only carbon isotopes with significant natural occurrence are $\ce{^{12}C}$ (98.9%) and $\ce{^{13}C}$ (1.1%). We can find all of the isotopic peaks and their relative abundances by then expanding the binomial $(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5$, where $m[^{12}C]$ and $m[^{13}C]$ denote the exact masses of the carbon-12 and carbon-13 isotopes, respectively. Expanding the binomial yields:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5 ={} & \ \ \ \ \ \binom {5} {0}(0.989\times m[^{12}C])^5 \\ & + \binom {5} {1}(0.989\times m[^{12}C])^4 \times (0.011\times m[^{13}C]) \\ & + \binom {5} {2}(0.989\times m[^{12}C])^3 \times (0.011\times m[^{13}C])^2 \\ & + \binom {5} {3}(0.989\times m[^{12}C])^2 \times (0.011\times m[^{13}C])^3 \\ & + \binom {5} {4}(0.989\times m[^{12}C]) \times (0.011\times m[^{13}C])^4 \\ & + \binom {5} {5} (0.011\times m[^{13}C])^5 \\ \end{aligned} \end{equation}$

Calculating the coefficients in each term:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times (m[^{13}C])^5 ={} & \ \ \ \ \ 0.946 \times (m[^{12}C])^5 \\ & + 0.0526\times (m[^{12}C])^4 \times m[^{13}C] \\ & + 0.00117 \times (m[^{12}C])^3 \times (m[^{13}C])^2 \\ & + 0.0000130 \times (m[^{12}C])^2 \times (m[^{13}C])^3 \\ & + 7.24×10^{-8}\times (m[^{12}C]) \times ( m[^{13}C])^4 \\ & + 1.61×10^{-10} \times m[^{13}C])^5 \\ \end{aligned} \end{equation}$

From the expansion, we see that 94.6% of all $\ce{C_5}$ molecules contain only carbon-12 (the lowest possible mass for the molecule), and almost all of the rest (5.3% out of the remaining 5.4%) is accounted for by molecules that contain a single carbon-13 atom. Only about 0.1% of $\ce{C_5}$ molecules contain two or more carbon-13 atoms.

But what happens in very large molecules? Intuitively, if there are many atoms, you would expect a higher chance of there being at least one less common isotope in the mix. Let's see the first few terms for the molecule $\ce{C_100}$:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ \binom {100} {0}(0.989\times m[^{12}C])^{100} \\ & + \binom {100} {1}(0.989\times m[^{12}C])^{99} \times (0.011\times m[^{13}C]) \\ & + \binom {100} {2}(0.989\times m[^{12}C])^{98} \times (0.011\times m[^{13}C])^2 \\ & + \ ...\\ \end{aligned} \end{equation}$

Calculating the coefficients:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ 0.331 \times (m[^{12}C])^{100} \\ & + 0.368 \times (m[^{12}C])^{99} \times (m[^{13}C]) \\ & + 0.203 \times (m[^{12}C])^{98} \times (m[^{13}C])^2 \\ & +\ ...\\ \end{aligned} \end{equation}$

Well that's interesting. Now only 33.1% of the molecules contain only carbon-12 atoms, and in fact more molecules contain exactly one carbon-13 atom, at 36.8% of the total. Even molecules with two carbon-13 atoms are quite abundant, at 20.3%.

Indeed, peaks containing rarer isotopes eventually dominate. For the huge molecule $\ce{C_10000}$, the strongest mass spectrum signal would come from molecules contaning 110 carbon-13 atoms, corresponding to 3.8% of the total, while a measly $9.2\times 10^{-47}\%$ of molecules contain only carbon-12. This happens because when $n$ is large, the term $\binom {n} {k}$ grows very quickly as $k$ rises from zero, overwhelming the increase in the exponent of the rarer isotope. You can see this behaviour quite nicely in this sequence of mass spectra of molecules with increasing size.

To calculate the specific $M/M+2$ ratio for a molecule containing only $n$ carbon atoms, all you need is to get the ratio for first and third terms in the binomial:

$\begin{equation} \begin{aligned} (0.989\times m[^{12}C] + 0.011\times m[^{13}C])^n ={} & \ \ \ \ \ \color{#0000ff}{ \binom {n} {0}(0.989\times m[^{12}C])^n} \\ & + \binom {n} {1}(0.989\times m[^{12}C])^{n-1} \times (0.011\times m[^{13}C]) \\ & + \color{#0000ff}{\binom {n} {2}(0.989\times m[^{12}C])^{n-2} \times (0.011\times m[^{13}C])^2} \\ & +\ ...\\ \end{aligned} \end{equation}$

The ratio is then:

$$\frac{\binom {n} {0}0.989^n}{\binom {n} {2}0.989^{n-2} \times 0.011^2}=\frac{2\times 0.989^2}{n(n-1)\times 0.011^2}$$

Technically this only holds if there are no other elements which contain multiple isotopes, though it will hold approximately if the other elements only have very rare alternate isotopes, such as hydrogen (99.98% hydrogen-1, 0.02% hydrogen-2).

As a last curiosity, all of the above extends to analysing more complicated molecules. For example, glucose ($\ce{C6H12O6}$) will have a mass spectrum exactly described by the expression:

$$(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^6 \times (0.9998\times m[^{1}H] + 0.002\times m[^{2}H])^{12} \times (0.9976\times m[^{16}O] + 0.004\times m[^{17}O] + 0.020\times m[^{18}O])^6$$

Happy expanding!