Calculating the Probability of Typographical Errors in a Manuscript: A Poisson Distribution Approach

When a manuscript of 400 pages contains 100 typographical errors, the question arises: what is the probability that a given page will contain exactly two errors?

Understanding the Poisson Distribution

The Poisson distribution is a statistical tool that is used to model the number of events occurring within a fixed interval. In this case, the interval is a single page in the manuscript, and the event is a typographical error. The Poisson distribution is suitable for this scenario because it assumes that the events (errors) occur independently and at a constant average rate.

Step-by-Step Calculation

Let's break down the process into several steps to find the probability that a given page contains exactly two typographical errors.

Step 1: Calculate the Average Number of Errors per Page

The average number of errors per page, denoted as lambda, can be calculated using the total number of errors and the total number of pages.

Formula: lambda frac{100}{400} 0.25

Step 2: Use the Poisson Probability Mass Function

The probability of observing exactly k errors in a given page can be calculated using the Poisson probability mass function (PMF).

Formula: PX k frac{lambda^k e^{-lambda}}{k!}

Where:
- X is the random variable representing the number of errors.
- k is the number of errors we want (2 in this case).
- lambda is the average number of errors per page (0.25 in this case).
- e is approximately equal to 2.71828.

Step 3: Plug in the Values

Now we can substitute the values into the formula to calculate the probability of finding exactly two errors on a given page.

PX 2 frac{0.25^2 e^{-0.25}}{2!}

0.25^2 0.0625 e^{-0.25} ≈ 0.7788 (using a calculator) 2! 2

Substituting these values back into the formula:

PX 2 frac{0.0625 times 0.7788}{2} frac{0.048675}{2} ≈ 0.0243375

Therefore, the probability that a given page contains exactly two typographical errors is approximately 0.0243 or 2.43%.

Alternative Approach Using Binomial Distribution

An alternative approach to this problem is using the binomial distribution. Here, we assume that there are 200 misprints and each one is placed randomly on a page. If we take a page at random and consider the 200 misprints being distributed randomly, the chance of a page receiving exactly 2 misprints can be calculated.

Formula: C(1, 200, 2) .995^198 .005^2 ≈ 0.1844

This result is based on the binomial distribution, where the probability of a page receiving exactly k successes (errors) out of n trials (total errors) is given by:

P(X k) C(n, k) * p^k * (1-p)^(n-k)

Where:

p 0.5 (average number of misprints per page) n 200 k 2

Conclusion

Both the Poisson and binomial distributions can be used to model the probability of typographical errors in a manuscript. The Poisson distribution is particularly useful when the number of pages and errors is large, making the assumption of constant rates more feasible. However, the binomial distribution provides a more detailed approach when the exact number of events (errors) and trials (pages) is known.