Question

A life insurance salesman sells on the average 3 life insurance policies per week. Assume that the number of policies per week follows the Poisson distribution and to calculate the probability that in a given week he will sell two or more policies. Round your answer to the nearest thousandth

Answer 1

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:

\[
\boxed{0.801}
\]