Testing for top 3 options I have data as shown in the chart below. There are 1144 rows of data and for each data row, an event (A, B, C, ..., I) could either have occurred or not occurred (that is, on or off; binary). For example, A and B could be present in one row and only C in another row. (The events actually represent characteristics that are present or not present in a user's drawing on a picture).
My hypothesis is that A, B, and C are the top 3 events (that is, occur the most often). Can I use Pearson's Chi-squared test to check this? I'm thinking that I cannot because the wiki page for it says, "The events considered must be mutually exclusive and have total probability 1." But in my case, the "events" are not mutually exclusive.
I'm currently using R with my data, but I'm a beginner to using it, so solutions with Excel or something else would also be appreciated.

The actual number of observations are as follows:
A   B   C   D   E   F   G   H   I
467 483 127 30  76  47  34  32  20

 A: Here is an approach that seems natural to me.
Introduction
Suppose there are $n$ observations of the following form: 
\begin{equation}
y_i = (y_{i1}, \ldots, y_{iJ}),
\end{equation}
where $y_{ij} \in \{0,1\}$. Let
\begin{equation}
Y = (Y_1, \ldots, Y_J),
\end{equation}
where 
\begin{equation}
Y_j = \sum_{i=1}^n y_{ij}.
\end{equation}
The problem
Consider the collection of subsets of $\{1,\ldots,J\}$, each containing $K < J$ elements. 
Let $(S_1, \ldots, S_L)$ denote the collection of those subsets, where $L = \binom{J}{K}$.
Let $M_\ell$ denote the sum of the elements in $\{Y_j: j\in S_\ell\}$:
\begin{equation}
M_\ell = \sum_{j\in S_\ell} Y_j.
\end{equation}
We wish to compute the probability that $M_\ell$ is the maximum:
\begin{equation}
\text{Pr}[\max(M_1, \ldots, M_L) = M_\ell] \qquad\text{for $\ell = 1, \ldots, L$}.
\end{equation}
A Bayesian framework
Think of $Y$ as being generated from a multinomial distribution:
\begin{equation}
p(Y|\theta) \propto \prod_{j=1}^J \theta_j^{Y_j},
\end{equation}
where $\theta = (\theta_1, \ldots, \theta_J) \in \Delta^{J-1}$, where $\Delta^{J-1}$ denotes the $J-1$-dimensional simplex. (In other words, $\theta_j \ge 0$ and $\sum_{j=1}^J \theta_j = 1$.) This provides a likelihood for $\theta$. Let the prior distribution for $\theta$ be given by the Dirichlet distribution:
\begin{equation}
p(\theta) = \textsf{Dirichlet}(\theta|\alpha) \propto \prod_{j=1}^J \theta_j^{\alpha_j-1},
\end{equation}
where $\alpha = (\alpha_1, \ldots, \alpha_J)$ and $\alpha_j > 0$. The mean of this distribution is $\alpha/\alpha_0$, where $\alpha_0 = \sum_{j=1}^J \alpha_j$. (The variance is inversely related to $\alpha_0$, which is called the concentration parameter.) Note that if $\alpha_j = 1$ for all $j$, then $p(\theta) \propto 1$, a flat distribution. 
The Dirichlet distribution is a conjugate prior for the multinomial likelihood. The posterior distribution for $\theta$ is given by
\begin{equation}
p(\theta|Y) \propto p(Y|\theta)\,p(\theta) \propto \prod_{j=1}^J \theta_j^{(\alpha_j + Y_j) - 1} \propto \textsf{Dirichlet}(\theta|\alpha'),
\end{equation}
where $\alpha' = \alpha + Y$. The posterior mean for $\theta$ is 
\begin{equation}
E[\theta|Y] = \alpha'/\alpha_0',
\end{equation}
where $\alpha_0' = \sum_{j=1}^J (\alpha_j + Y_j)$.
Solution to the problem
The posterior distribution for $\theta$ summarizes what is known about the categories $(1, \ldots, J)$. With this in mind, switch perspective from the elements of $Y$ to the corresponding elements of $\theta$ and use $p(\theta|Y)$ to solve the problem posed above. 
Let $m_\ell$ denote the sum of the elements in $\{\theta_j : \theta_j \in S_\ell\}$: 
\begin{equation}
m_\ell = \sum_{j\in S_\ell} \theta_j.
\end{equation}
The joint posterior distribution for $(m_1, \ldots, m_L)$ can be computed from the posterior distribution for $\theta$. 
The posterior mean is one feature of the distribution,
\begin{equation}
E[m_\ell|Y] = \sum_{j \in S_\ell} E[\theta_j|Y] = \frac{1}{\alpha_0'} \sum_{j\in S_\ell} \alpha_j'.
\end{equation}
We wish to compute the posterior probability that $m_\ell$ is the maximal element:
\begin{equation}
\pi_\ell = \text{Pr}[\max(m_1, \ldots, m_L) = m_\ell\,|\,Y] \qquad\text{for $\ell = 1, \ldots, L$}.
\end{equation}
These probabilities will depend on the variance of $p(\theta|Y)$ as well as its mean.
An analytical solution may be possible, but here I describe a numerical solution that is simple to implement. 
Make $R$ draws from $\textsf{Dirichelt}(\theta|\alpha')$ and let $r \in \{1,\ldots, R\}$ index the draws. For each draw compute 
\begin{equation}
z^{(r)} = \arg\!\max_{\ell}\ (m_1^{(r)}, \ldots, m_L^{(r)}).
\end{equation} 
Note $z^{(r)} \in \{1, \ldots, L\}$.
Then 
\begin{equation}
\pi_\ell \approx \frac{1}{R} \sum_{r=1}^R [z^{(r)} = \ell],
\end{equation}
where $[\,\cdot\,]$ is the Iverson bracket defined as
\begin{equation}
[x] = \begin{cases}
1 & \text{$x$ is true} \\
0 & \text{$x$ is false}
\end{cases} .
\end{equation}
The larger $R$ is, the better the approximation. (One can compute a numerical standard error to assess the accuracy of the approximation.)
Example
Let $J = 9$ and 
\begin{equation}
Y = (467,\ 483,\ 127,\ 30,\  76,\  47,\  34,\  32,\  20).
\end{equation}
In addition, let $\alpha_j = 1$ for all $j \in \{1, \ldots, 9\}$. (Note $\alpha_0' = 1325$.) The mean of the posterior distribution is 
\begin{equation}
E[\theta|Y] \approx (\text{0.353},\text{ 0.365},\text{ 0.097},\text{
   0.023},\text{ 0.058},\text{ 0.036},\text{
   0.026},\text{ 0.025},\text{ 0.016}).
\end{equation}
Let $K = 3$, in which case $L = 84$. Let $S_1 = \{1,2,3\}$ and $S_3 = \{1,2,5\}$. Then $E[m_1|Y] \approx 0.815$ and $E[m_3|Y] \approx 0.777$.
In $R = 10^6$ draws, only $m_1$ and $m_3$ appeared as the maximum, with $m_3$ appearing 170 times. Therefore $\pi_1 \approx 0.998$, $\pi_3 \approx 0.002$, and $\pi_\ell \approx 0$ for all other $\ell$.
