diff --git a/00014_Medical_tests/en_article_proofreading.md b/00014_Medical_tests/en_article_proofreading.md
index 8519e40..1b4f85d 100644
--- a/00014_Medical_tests/en_article_proofreading.md
+++ b/00014_Medical_tests/en_article_proofreading.md
@@ -1,129 +1,127 @@
----
-keywords:
-- conditional probability
-- frequency of occurrence
-is_finished: False
----
-
-# Diagnostic tests in medicine
-
-Every person undergoes many diagnostic tests during their lifetime
-either at the doctor's office or at home. And it doesn't matter
-whether it is a test for COVID-19, celiac disease,
-general mammogram screening or home pregnancy test.
-In any case, we want to know whether the result of a test
-is reliable. 
-For each such test, two basic data are given.
-
-- *test sensitivity*, i.e. the probability that the test will detect the disease when I have it;
-- *specificity of the test*, i.e. the probability that the test is negative when I do not have the disease.
-
-So-called *prevalence of the disease*, i.e. the ratio of the number of individuals who have the disease to the number of all individuals in the population, is also important for the evaluation.
-
-## Test result interpretation
-
-A number of home tests for allergies and intolerances are sold in pharmacies and drugstores, 
-but also COVID-19 or pregnancy tests. 
-For example on the leaflet of the test for celiac disease (gluten intolerance), 
-we can read that the sensitivity of the test is $96{,}3\,\%$ 
-and the specificity of the test is $89{,}7\,\%$. We know that, 
-that the number of celiacs in the population is roughly $1\,\%$, so the prevalence of the disease is $1/100$.
-
-We will be primarily interested in the question: If I test positive, 
-what is the likelihood that I have celiac disease? Try to guess the answer first. 
-In the following exercise, we will make an exact calculation.
-
-> **Exercise 1.** The test sensitivity is $96{,}3\,\%$, the test specificity is $89{,}7\,\%$
-> and the prevalence of the disease is $1/100$. What is the probability
-> that a patient who tests positive for celiac disease is actually sick?
-
-*Solution.* This is an example of conditional probability 
-that can be easily solved using Bayes' theorem:
-
-Let $A$ be a random phenomenon and let $B_1,\dots, B_n$ be a complete system of phenomena. 
-If $P(A)>0$, $P(B_1)>0,\dots,P(B_n)>0$, then
-$$
-P(B_k|A)=\frac{P(A\cap B_k)}{P(A)}=\frac{P(A|B_k)P(B_k)}{\sum_{i=1}^{n}P(A|B_i)P(B_i)},\quad k=1,\dots,n.
-$$
-
-Symbol $P(B|A)$ denotes the probability that event $B$ occurred when event $A$ occurred.
-
-Let's denote
-
- - $P(C)$ the probability that I'm positive,
- - $P(N)$ the probability that I'm negative,
- - $P(+|C)$ the probability that the test is positive if I'm positive,
- - $P(-|C)$ the probability that the test is negative when I am positive,
- - $P(+|N)$ the probability that the test is positive when I'm negative,
- - $P(-|N)$ the probability that the test is negative when I'm negative,
- - $P(C|+)$ the probability that I am positive when the test is positive.
-  
-Since the prevalence is $1/100$, then
-$$
-P(C)=0{,}01 \qquad P(N)=0{,}99
-$$
-and from the assignment 
-$$
-P(+|C)=0{,}963 \qquad  P(-|C)=0{,}037 \qquad P(+|N)=0{,}103 \qquad P(-|N)=0{,}897.
-$$
-
-According to Bayes' theorem, it holds 
-
-$$
-\begin{aligned}
-P(C|+)&=\frac{P(+|C)P(C)}{P(+|C)P(C)+P(+|N)P(N)}=\\
-&=\frac{0{,}963\cdot0{,}01}{0{,}963\cdot0{,}01+0{,}103\cdot0{,}99}=0{,}086.
-\end{aligned}
-$$
-
-The probability that a positive tested patient is really sick 
-is therefore only $0{,}086$, i.e. $8{,}6\,\%$ .
-
-It is very likely that many students will not guess this result before the calculation.
-This is not surprising, as studies have shown that up to 95% 
-of doctors cannot answer questions like this correctly.
-
-> **Exercise 2.** What is the probability that a patient who tests negative is really negative?
-
-*Solution.* We will use the notation from the previous exercise and Bayes' theorem. We get 
-$$
-\begin{aligned}
-P(N|-)&=\frac{P(-|N)P(N)}{P(-|N)P(N)+P(-|C)P(C)}=\\
-&=\frac{0{,}897\cdot0{,}99}{0{,}897\cdot0{,}99+0{,}037\cdot0{,}01}=0{,}999583.
-\end{aligned}
-$$
-The probability that a negatively tested patient is actually healthy is therefore
-$0{,}999583$, i.e. $99{,}96\,\%$ .
-
-
-## Interpretation without using probability
-
-Let's try to solve the first exercise again, but we will change the assignment to avoid the concept of conditional probability.
-
-> **Exercise 3.** Out of 100 000 people, about 1 000 have celiac disease.
->Of these 1 000 people, the test detects celiac disease in 960 cases.
->Of the other 99 000 people who don't have celiac disease,
->10 200 people will also test positive.
->Let's imagine a sample of people who test positive.
->What proportion of them actually have celiac disease?
-
-*Solution.* In total we have $$960+10\,200=11\,160$$  
-positive results, but only $960$ really sick ones, i.e.
-
-$$ \frac{960}{11\,160}=0,086 \quad\Longrightarrow\quad 8{,}6\,\%. $$
-We have made two simplifications in this calculation. 
-Instead of working with decimal numbers, we worked with integers rounded to tens. 
-Thus, we have sacrificed a little precision for simplicity. 
-In addition, we assumed that everything would turn out as probability predicts, 
-i.e., for example, that exactly 1 000 people have celiac disease. 
-This may not always be true. We don't usually find this procedure in textbooks,
-but people usually understand it better. With this exercise statement, 
-the success rate of doctors in estimating the outcome is almost 100%.
-
-## Literature
-
-1. Singh P., Arora A., Strand T.A., Leffler D.A., Catassi C., Green P.H., Kelly C.P., Ahuja V., Makharia G.K., 
-   *Global Prevalence of Celiac Disease: Systematic Review and Meta-analysis*.
-   Clin Gastroenterol Hepatol. 2018 Jun; 16(6):823-836. doi: 10.1016/j.cgh.2017.06.037.
-2. Gigerenzer, G., *Calculated risks*, Simon and Schuster, 2003. 
-
+---
+keywords:
+- conditional probability
+- frequency of occurrence
+is_finished: False
+---
+
+# Diagnostic Tests in Medicine
+
+Every person undergoes many diagnostic tests during their lifetime,
+either at the doctor's office or at home. It doesn't matter
+whether it is a test for COVID-19, celiac disease,
+general mammogram screening, or a home pregnancy test.
+In any case, we want to know whether the test result
+is reliable. 
+For each such test, two basic pieces of information are provided:
+
+- *Test Sensitivity*: the probability that the test will detect the disease when I have it.
+- *Test Specificity*: the probability that the test will be negative when I do not have the disease.
+
+The so-called *prevalence of the disease*, i.e., the ratio of the number of individuals who have the disease to the number of all individuals in the population, is also important for evaluation.
+
+## Test Result Interpretation
+
+A number of home tests for allergies and intolerances are sold in pharmacies and drugstores, 
+including COVID-19 or pregnancy tests. 
+For example, the leaflet of a test for celiac disease (gluten intolerance) states that the sensitivity of the test is $96{,}3\,\%$ 
+and the specificity of the test is $89{,}7\,\%$. We know that the number of celiacs in the population is roughly $1\,\%$, so the prevalence of the disease is $1/100$.
+
+We will be primarily interested in the question: If I test positive, 
+what is the probability that I actually have celiac disease? Try to guess the answer first. 
+In the following exercise, we will perform the exact calculation.
+
+> **Exercise 1.** The test sensitivity is $96{,}3\,\%$, the test specificity is $89{,}7\,\%$
+> and the prevalence of the disease is $1/100$. What is the probability
+> that a patient who tests positive for celiac disease actually has the disease?
+
+*Solution.* This is an example of a problem on conditional probability
+that can be easily solved using Bayes' theorem:
+
+Let $A$ be a random event and let $B_1,\dots, B_n$ be a complete set of events. <!-- termín “phenomena” se v tomto kontextu moc nepoužívá. Písmenka ve vzorci zastupují “jevy” čemuž odpovídá běžně používaný termín “events”. -->
+If $P(A)>0$, $P(B_1)>0,\dots,P(B_n)>0$, then
+$$
+P(B_k|A)=\frac{P(A\cap B_k)}{P(A)}=\frac{P(A|B_k)P(B_k)}{\sum_{i=1}^{n}P(A|B_i)P(B_i)},\quad k=1,\dots,n.
+$$
+
+The symbol $P(B|A)$ denotes the probability that event $B$ occurs given that event $A$ has occurred.
+
+Let's define the following probabilities:
+
+ - $P(C)$ the probability that I'm positive (I have the disease),
+ - $P(N)$ the probability that I'm negative (I don’t have the disease),
+ - $P(+|C)$ the probability that the test is positive if I'm positive,
+ - $P(-|C)$ the probability that the test is negative if I’m positive,
+ - $P(+|N)$ the probability that the test is positive  if I'm negative,
+ - $P(-|N)$ the probability that the test is negative if I'm negative,
+ - $P(C|+)$ the probability that I’m positive if the test is positive.
+  
+Since the prevalence is $1/100$, then
+$$
+P(C)=0{,}01 \qquad P(N)=0{,}99
+$$
+and from the given data 
+$$
+P(+|C)=0{,}963 \qquad  P(-|C)=0{,}037 \qquad P(+|N)=0{,}103 \qquad P(-|N)=0{,}897.
+$$
+
+According to Bayes' theorem, it holds that
+
+$$
+\begin{aligned}
+P(C|+)&=\frac{P(+|C)P(C)}{P(+|C)P(C)+P(+|N)P(N)}=\\
+&=\frac{0{,}963\cdot0{,}01}{0{,}963\cdot0{,}01+0{,}103\cdot0{,}99}=0{,}086.
+\end{aligned}
+$$
+
+The probability that a  patient who tests positive really has the disease 
+is therefore only $0{,}086$, i.e., $8{,}6\,\%$ .
+
+*Note.* It is very likely that many students will not guess this result correctly before the calculation.
+This is not surprising, as studies have shown that up to 95% 
+of doctors cannot answer questions like this correctly.
+
+> **Exercise 2.** What is the probability that a patient who tests negative is really negative?
+
+*Solution.* We will use the notation from the previous exercise and Bayes' theorem. We get 
+$$
+\begin{aligned}
+P(N|-)&=\frac{P(-|N)P(N)}{P(-|N)P(N)+P(-|C)P(C)}=\\
+&=\frac{0{,}897\cdot0{,}99}{0{,}897\cdot0{,}99+0{,}037\cdot0{,}01}=0{,}999583.
+\end{aligned}
+$$
+Thus, the probability that  a patient who tests negative is actually healthy is
+$0{,}999583$, i.e., $99{,}96\,\%$.
+
+
+## Interpretation Without Using Probability
+
+Let's try to solve the first exercise again, but we will reframe the assignment to avoid the concept of conditional probability.
+
+> **Exercise 3.** Out of 100 000 people, about 1 000 have celiac disease.
+>Of these 1 000 people, the test detects celiac disease in 960 cases.
+>Of the remining 99 000 people who don't have celiac disease,
+>10 200 people will also test positive.
+>Let's imagine the group of those people who test positive.
+>What proportion of them actually have celiac disease?
+
+*Solution.* In total, we have $$960+10\,200=11\,160$$  
+positive test results, but only $960$ are true positive cases, so:
+
+$$ \frac{960}{11\,160}=0,086 \quad\Longrightarrow\quad 8{,}6\,\%. $$
+<!-- Tady bych doporučovala přidat ještě slovní odpověď na slovní zadání úlohy. Třeba:--> 
+We see again that the proportion of people who have the disease among those who tested positive is $8{,}6\, \%$.  
+
+In this calculation, we simplified by working with whole numbers and rounding to the nearest ten. <!-- Tady navrhuji trochu přeformulovat, protože se v první větičce mluví (myslím zbytečně) o dvou zjednodušeních, ale ta, za mně, nejsou jasně oddělena.-->  Thus, we have sacrificed a little precision for simplicity. 
+In addition, we assumed that everything would turn out as probability predicts, 
+i.e.,  that exactly 1 000 people have celiac disease. 
+This is not always true. Although this method is simpler and often easier to understand, it is not typically found in textbooks. However, with this exercise statement, 
+the success rate of doctors in estimating the result is almost 100%.
+
+## Literature
+
+1. Singh P., Arora A., Strand T.A., Leffler D.A., Catassi C., Green P.H., Kelly C.P., Ahuja V., Makharia G.K., 
+   *Global Prevalence of Celiac Disease: Systematic Review and Meta-analysis*.
+   Clin Gastroenterol Hepatol. 2018 Jun; 16(6):823-836. doi: 10.1016/j.cgh.2017.06.037.
+2. Gigerenzer, G., *Calculated risks*, Simon and Schuster, 2003. 
+