Mr. A writes an article. The article originally is error free. Each day Mr. B introduces one new error into the article. At the end of the day, Mr. A checks the article and has `(2)/(3)` chance of catching each individual error still in the article. After 3 days, the probability that the article is error free can be expressed as `(p)/(q)` where p and q are relatively prime positive integers. Let `lambda=q-p`, then find the sum of the digits of `lambda`.

Answer

To solve the problem step by step, we need to calculate the probability that the article is error-free after 3 days, given that Mr. B introduces one error each day and Mr. A has a \( \frac{2}{3} \) chance of catching each error.

### Step 1: Understanding the Errors Introduced
- Each day, Mr. B introduces one new error into the article.
- After 3 days, there will be a total of 3 errors in the article (one error introduced each day).

### Step 2: Probability of Catching Errors
- Mr. A has a \( \frac{2}{3} \) chance of catching each error.
- Therefore, the probability of not catching an error is \( 1 - \frac{2}{3} = \frac{1}{3} \).

### Step 3: Probability of Not Catching All Errors
- The probability that Mr. A does not catch any specific error is \( \frac{1}{3} \).
- Since there are 3 errors, the probability that Mr. A does not catch any of the 3 errors is:
\[ \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

### Step 4: Probability of Catching at Least One Error
- The probability that Mr. A catches at least one error (which means the article is not error-free) is:
\[ 1 - \text{Probability of not catching any errors} = 1 - \frac{1}{27} = \frac{26}{27} \]

### Step 5: Probability of the Article Being Error-Free
- The probability that the article is error-free after 3 days is the same as the probability that Mr. A catches all errors:
\[ \text{Probability of being error-free} = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \]

### Step 6: Expressing the Probability in Terms of \( p \) and \( q \)
- Here, the probability \( \frac{8}{27} \) can be expressed as \( \frac{p}{q} \) where \( p = 8 \) and \( q = 27 \).
- Since 8 and 27 are relatively prime, we can proceed to the next step.

### Step 7: Calculating \( \lambda \)
- We define \( \lambda = q - p \):
\[ \lambda = 27 - 8 = 19 \]

### Step 8: Finding the Sum of the Digits of \( \lambda \)
- The sum of the digits of \( \lambda = 19 \) is:
\[ 1 + 9 = 10 \]

### Final Answer
The sum of the digits of \( \lambda \) is \( 10 \).

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