Update evaluation.mdx

According to the definition given of Insertion and deletion operation, changed

chapters/en/chapter5/evaluation.mdx

@@ -36,20 +36,20 @@ insertions and deletions on the *word level*. This means errors are annotated on
 | Reference: | the | cat | sat | on | the | mat |
 |-------------|-----|-----|---------|-----|-----|-----|
 | Prediction: | the | cat | **sit** | on | the | | |
-| Label: | ✅ | ✅ | S | ✅ | ✅ | D |
+| Label: | ✅ | ✅ | S | ✅ | ✅ | I |
 
 Here, we have:
 * 1 substitution ("sit" instead of "sat")
-* 0 insertions
-* 1 deletion ("mat" is missing)
+* 1 insertions ("mat" is missing)
+* 0 deletion 
 
 This gives 2 errors in total. To get our error rate, we divide the number of errors by the total number of words in our
 reference (N), which for this example is 6:
 
 $$
 \begin{aligned}
 WER &= \frac{S + I + D}{N} \\
-&= \frac{1 + 0 + 1}{6} \\
+&= \frac{1 + 1 + 0}{6} \\
 &= 0.333
 \end{aligned}
 $$
@@ -116,17 +116,17 @@ individual characters, and annotate errors on a character-by-character basis:
 | Reference: | t | h | e | | c | a | t | | s | a | t | | o | n | | t | h | e | | m | a | t |
 |-------------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|
 | Prediction: | t | h | e | | c | a | t | | s | **i** | t | | o | n | | t | h | e | | | | |
-| Label: | ✅ | ✅ | ✅ | | ✅ | ✅ | ✅ | | ✅ | S | ✅ | | ✅ | ✅ | | ✅ | ✅ | ✅ | | D | D | D |
+| Label: | ✅ | ✅ | ✅ | | ✅ | ✅ | ✅ | | ✅ | S | ✅ | | ✅ | ✅ | | ✅ | ✅ | ✅ | | I | I | I |
 
 We can see now that for the word "sit", the "s" and "t" are marked as correct. It's only the "i" which is labelled as a
 substitution error (S). Thus, we reward our system for the partially correct prediction 🤝
 
-In our example, we have 1 character substitution, 0 insertions, and 3 deletions. In total, we have 14 characters. So, our CER is:
+In our example, we have 1 character substitution, 3 insertions, and 0 deletions. In total, we have 14 characters. So, our CER is:
 
 $$
 \begin{aligned}
 CER &= \frac{S + I + D}{N} \\
-&= \frac{1 + 0 + 3}{14} \\
+&= \frac{1 + 3 + 0}{14} \\
 &= 0.286
 \end{aligned}
 $$