To solve the problem of finding the probability that an author wrote at least one check on a randomly selected day using the Poisson distribution, follow these steps:
1. Identify the given information:
- The total number of checks written in a year is 193.
- There are 365 days in a year.
2. Calculate the average number of checks written per day (λ):
- The average (λ) can be calculated as the total number of checks divided by the number of days in the year:
[tex]\[
\lambda = \frac{193 \text{ checks}}{365 \text{ days}} \approx 0.528767 \text{ checks per day}
\][/tex]
3. Understand the Poisson distribution:
- The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space.
- The probability of observing [tex]\( k \)[/tex] events in an interval is given by:
[tex]\[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
\][/tex]
4. Calculate the probability of writing no checks on a randomly selected day (k = 0):
- For [tex]\( k = 0 \)[/tex]:
[tex]\[
P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-\lambda}
\][/tex]
- Using the average number of checks per day ([tex]\( \lambda \approx 0.528767 \)[/tex]):
[tex]\[
P(X = 0) \approx e^{-0.528767} \approx 0.589331
\][/tex]
5. Calculate the probability of writing at least one check:
- The probability of writing at least one check is the complement of the probability of writing zero checks:
[tex]\[
P(X \geq 1) = 1 - P(X = 0)
\][/tex]
- Using the previously calculated probability of writing no checks:
[tex]\[
P(X \geq 1) = 1 - 0.589331 \approx 0.411
\][/tex]
6. Round the answer to three decimal places:
- The probability that the author wrote at least one check on a randomly selected day is:
[tex]\[
\boxed{0.411}
\][/tex]
Therefore, the probability that the author wrote at least one check on a randomly selected day is 0.411.
