Give step-by-step solution with explanation and final answer:20.pogn manu=crf has 3 ypos ha ae see random 370s ho po
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ACE iii - 7
3 mrt to pr cy op os 2m? 7
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| Wha orbit md hor sac cr eo st 400057
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Wht 5 cy cy cn ft 0 pe?
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CATR T———
Wha erty kor re ach 12 pos md £0 ges?
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Question:
Give step-by-step solution with explanation and final answer:20.pogn manu=crf has 3 ypos ha ae see random 370s ho po
es Cf lt i Cc
ACE iii - 7
3 mrt to pr cy op os 2m? 7
nl
| Wha orbit md hor sac cr eo st 400057
hed sc en
Wht 5 cy cy cn ft 0 pe?
[un]
CATR T———
Wha erty kor re ach 12 pos md £0 ges?
0
Foundiofous docm laces a ect)
This is AI generated Response.
Asked by: anil
Created at: 2025-09-29 07:35:56
Final Answer:
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Let's break down and solve each part step-by-step.
### **Given:**
- 500-page manuscript
- 30 typos in total
- Typos follow a **Poisson distribution**
- **Average rate of typos per page:**
\[
\lambda = \frac{30 \text{ typos}}{500 \text{ pages}} = .06 \text{ typos/page}
\]
- For \( n \) pages:
\[
\text{Expected number of typos} = \lambda' = n \times .06
\]
The **Poisson probability formula**:
\[
P(X = k) = \frac{e^{-\lambda'} \cdot (\lambda')^k}{k!}
\]
Where:
- \( \lambda' \) = expected number of typos in the interval,
- \( k \) = exact number of typos,
- \( e \) ≈ 2.71828.
---
## **a. Probability of exactly 1 typo in the first 20 pages**
- \( n = 20 \)
- \( \lambda' = 20 \times .06 = 1.2 \)
- \( k = 1 \)
\[
P(X = 1) = \frac{e^{-1.2} \cdot 1.2^1}{1!}
\]
\[
e^{-1.2} \approx .3012
\]
\[
P(X=1) = .3012 \times 1.2 = .3614
\]
**Final Answer (a):**
\[
\boxed{.3614}
\]
---
## **b. Probability of exactly 1 typo in the first 40 pages**
- \( n = 40 \)
- \( \lambda' = 40 \times .06 = 2.4 \)
- \( k = 1 \)
\[
P(X = 1) = \frac{e^{-2.4} \cdot 2.4^1}{1!}
\]
\[
e^{-2.4} \approx .0907
\]
\[
P(X=1) = .0907 \times 2.4 = .2177
\]
**Final Answer (b):**
\[
\boxed{.2177}
\]
---
## **c. Probability of exactly 6 typos in the first 80 pages**
- \( n = 80 \)
- \( \lambda' = 80 \times .06 = 4.8 \)
- \( k = 6 \)
\[
P(X = 6) = \frac{e^{-4.8} \cdot 4.8^6}{6!}
\]
\[
e^{-4.8} \approx .0082
\]
\[
4.8^6 = 12288.
\]
\[
6! = 720
\]
\[
P(X=6) = .0082 \times \frac{12288}{720} = .0082 \times 17.0667 = .140
\]
**Final Answer (c):**
\[
\boxed{.140}
\]
---
## **d. Probability of exactly 12 typos in the first 140 pages**
- \( n = 140 \)
- \( \lambda' = 140 \times .06 = 8.4 \)
- \( k = 12 \)
\[
P(X = 12) = \frac{e^{-8.4} \cdot 8.4^{12}}{12!}
\]
\[
e^{-8.4} \approx .0002249
\]
\[
8.4^{12} \approx 1,155,929,882
\]
\[
12! = 479001600
\]
\[
P(X=12) = .0002249 \times \frac{1,155,929,882}{479,001,600}
\]
\[
\frac{1,155,929,882}{479,001,600} \approx 2.413
\]
\[
.0002249 \times 2.413 \approx .543
\]
**Final Answer (d):**
\[
\boxed{.543}
\]
---
### **Summary Table**
| Part | Probability (rounded to 4 decimals) |
|------|------------------------------------|
| a | .3614 |
| b | .2177 |
| c | .140 |
| d | .543 |
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