To find the probability that the salesman sells two or more life insurance policies in a week, we will use the properties of the Poisson distribution.
The Poisson distribution is defined by the formula:
\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]
where:
- \( \lambda \) is the average number of occurrences (in this case, 3 policies per week),
- \( k \) is the actual number of occurrences (policies sold),
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
We want to calculate \( P(X \geq 2) \), which can be found using the complementary probability:
\[ P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) \]
Step 1: Calculate \( P(X = 0) \)
\[ P(X = 0) = \frac{3^0 e^{-3}}{0!} = \frac{1 \cdot e^{-3}}{1} = e^{-3} \approx 0.0498 \]
Step 2: Calculate \( P(X = 1) \)
\[ P(X = 1) = \frac{3^1 e^{-3}}{1!} = \frac{3 \cdot e^{-3}}{1} = 3 e^{-3} \approx 3 \cdot 0.0498 \approx 0.1494 \]
Step 3: Combine to find \( P(X < 2) \)
\[ P(X < 2) = P(X = 0) + P(X = 1) \approx 0.0498 + 0.1494 = 0.1992 \]
Step 4: Calculate \( P(X \geq 2) \)
\[ P(X \geq 2) = 1 - P(X < 2) \approx 1 - 0.1992 = 0.8008 \]
Final Answer
Rounding to the nearest thousandth, the probability that the salesman will sell two or more policies in a given week is:
\[ \boxed{0.801} \]