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The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events (in this case, typographical errors) occurring in a fixed interval. The probability mass function of the Poisson distribution is given by: \[P(k;\lambda) = e^{-\lambda}\frac{\lambda^k}{k!}\] Where: - \(k\) is the number of events (typographical errors), - \(\lambda\) is the average number of events in the interval (the expected number of typographical errors per page, which is 0.2 in this case), - \(e\) is the base of the natural logarithm (approximately equal to 2.71828).

To find the probability that the next page we read contains 0 typographical errors, we will use the Poisson distribution formula with \(k = 0\): \(P(0; 0.2) = e^{-0.2}\frac{0.2^0}{0!}\) Since \(0! = 1\) and \(0.2^0 = 1\), then: \(P(0; 0.2) = e^{-0.2}\) Calculate the value of the probability: \(P(0; 0.2) \approx 0.8187\) Therefore, the probability that the next page we read contains 0 typographical errors is approximately 0.8187.

To find the probability that the next page we read contains 2 or more typographical errors, we will use the Poisson distribution formula and the complement rule. First, we need to find the probability of 1 or fewer errors (i.e., 0 or 1 errors), then subtract that probability from 1. Calculate the probability of 1 error: \(P(1; 0.2) = e^{-0.2}\frac{0.2^1}{1!}\) \(P(1; 0.2) = 0.2e^{-0.2}\) \(P(1; 0.2) \approx 0.1637\) Now, use the complement rule to get the probability of 2 or more errors: \(P(2 \text{ or more}; 0.2) = 1 - (P(0; 0.2) + P(1; 0.2))\) \(P(2 \text{ or more}; 0.2) \approx 1 - (0.8187 + 0.1637)\) \(P(2 \text{ or more}; 0.2) \approx 0.0176\) Therefore, the probability that the next page we read contains 2 or more typographical errors is approximately 0.0176.