2. (2 points) In 1898 L . J. Bortkiewicz published a book entitled The Law of Small Numbers. He used dat collected over 20 years to show that the number of soldiers killed by horse kicks each year in each corp in the Prussian cavalry followed a Poisson distribution with a mean of 0.5 . What is the probability of two or more deaths in a corps in three years? Answer:

To solve this problem, we need to calculate the probability of two or more deaths in a corps in three years, given that the number of deaths follows a Poisson distribution with a mean of 0.5 per year. ### Step 1: Determine the mean for three years Since the mean number of deaths per year is \( \lambda = 0.5 \), for three years, the mean will be: \[ \lambda_{3 \text{ years}} = 3 \times 0.5 = 1.5 \] ### Step 2: Use the Poisson probability formula The probability of observing \( k \) events (deaths) in a Poisson distribution is given by the formula: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( e \) is the base of the natural logarithm, \( \lambda \) is the average rate (mean), and \( k \) is the number of occurrences. ### Step 3: Calculate the probability of 0 and 1 death To find the probability of two or more deaths, we can first calculate the probabilities of 0 and 1 death, and then subtract these from 1. 1. **Probability of 0 deaths**: \[ P(X = 0) = \frac{e^{-1.5} (1.5)^0}{0!} = e^{-1.5} \] 2. **Probability of 1 death**: \[ P(X = 1) = \frac{e^{-1.5} (1.5)^1}{1!} = e^{-1.5} \cdot 1.5 \] ### Step 4: Calculate the total probability of 0 or 1 death \[ P(X < 2) = P(X = 0) + P(X = 1) = e^{-1.5} + e^{-1.5} \cdot 1.5 \] ### Step 5: Calculate the probability of 2 or more deaths \[ P(X \geq 2) = 1 - P(X < 2) \] Now, let's perform the calculations. First, we need to calculate \( e^{-1.5} \). Then we can find \( P(X < 2) \) and finally \( P(X \geq 2) \). Let's calculate these values. Calculate the value by following steps: - step0: Calculate: \(e^{-1.5}\) - step1: Convert the expressions: \(e^{-\frac{3}{2}}\) - step2: Express with a positive exponent: \(\frac{1}{e^{\frac{3}{2}}}\) - step3: Multiply by the Conjugate: \(\frac{e^{\frac{1}{2}}}{e^{\frac{3}{2}}\times e^{\frac{1}{2}}}\) - step4: Multiply the numbers: \(\frac{e^{\frac{1}{2}}}{e^{2}}\) Calculate or simplify the expression \( 0.22313 + (0.22313 * 1.5) \). Calculate the value by following steps: - step0: Calculate: \(0.22313+\left(0.22313\times 1.5\right)\) - step1: Multiply the numbers: \(0.22313+0.334695\) - step2: Add the numbers: \(0.557825\) Calculate or simplify the expression \( e^(-1.5) * 1.5 \). Calculate the value by following steps: - step0: Calculate: \(e^{-1.5}\times 1.5\) - step1: Convert the expressions: \(e^{-\frac{3}{2}}\times 1.5\) - step2: Convert the expressions: \(e^{-\frac{3}{2}}\times \frac{3}{2}\) - step3: Multiply the numbers: \(\frac{3}{2e^{\frac{3}{2}}}\) - step4: Multiply by the Conjugate: \(\frac{3e^{\frac{1}{2}}}{2e^{\frac{3}{2}}\times e^{\frac{1}{2}}}\) - step5: Multiply the numbers: \(\frac{3e^{\frac{1}{2}}}{2e^{2}}\) Calculate or simplify the expression \( 1 - 0.557825 \). Calculate the value by following steps: - step0: Calculate: \(1-0.557825\) - step1: Subtract the numbers: \(0.442175\) ### Step-by-Step Calculation Summary 1. **Mean for three years**: \[ \lambda_{3 \text{ years}} = 1.5 \] 2. **Probability of 0 deaths**: \[ P(X = 0) = e^{-1.5} \approx 0.22313 \] 3. **Probability of 1 death**: \[ P(X = 1) = e^{-1.5} \cdot 1.5 \approx 0.334695 \] 4. **Total probability of 0 or 1 death**: \[ P(X < 2) = P(X = 0) + P(X = 1) \approx 0.22313 + 0.334695 \approx 0.557825 \] 5. **Probability of 2 or more deaths**: \[ P(X \geq 2) = 1 - P(X < 2) \approx 1 - 0.557825 \approx 0.442175 \] ### Final Answer The probability of two or more deaths in a corps in three years is approximately \( \boxed{0.4422} \).