Hypothesis testing with binary pattern
Given what you explained, you can count the number of times the sequence is
001, 100 or 010. Then, $\hat{p}$ is defined as that count divided by the number of sequences, $N$.
If we define the null as the hypothesis that the proportion of good sequences  is 1.0,  then the test is
$H_0: p = 1.0$ versus $H_1: p \neq 1.0$
As far as testing the null, ( I'm assuming N is large here so that the normal
approximation to the binomial holds )  you can calculate the test statistic
as $Z =  \frac{ \hat{p} - p_{0}}{\sqrt{\frac{\bar{p}(1-\bar{p})}{N}}}$
and use $p_{0} = 1 $ and $\bar{p} = 0.5$. Then, reject the null if $|Z| &gt; Z_{\frac{\alpha}{2}}$ where $\alpha$ is the critical value of the test.
The CI  would be $\hat{p} ~ \pm Z_{\frac{\alpha}{2}} \times 
{\sqrt{\frac{\bar{p}(1-\bar{p})}{N}}}$.
If a digit in one of the three digit sequences is is missing or there's an extra, and you know this, then you can just count that sequence as a bad sequence.