An insurance company finds that .005 percent of the population die from a certain kind of accident each year. What is the probability that the company must pay off on more than 3 of 10,000 insured risks against such accidents in a given year?
The Correct Answer and Explanation is :
We are given that the probability of an individual dying from a certain kind of accident in a year is 0.005%, or:
[
p = \frac{0.005}{100} = 0.00005
]
For a total of 10,000 insured individuals, we define ( X ) as the number of deaths in a given year. ( X ) follows a Binomial distribution:
[
X \sim \text{Binomial}(n=10,000, p=0.00005)
]
Step 1: Check if Poisson Approximation is Valid
Since ( n ) is large and ( p ) is small, we approximate the Binomial distribution using a Poisson distribution with:
[
\lambda = np = 10,000 \times 0.00005 = 0.5
]
Thus, we approximate ( X ) as:
[
X \sim \text{Poisson}(\lambda = 0.5)
]
We need to find the probability that more than 3 deaths occur:
[
P(X > 3) = 1 – P(X \leq 3)
]
Step 2: Compute Cumulative Probability ( P(X \leq 3) )
Using the Poisson probability mass function:
[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
]
we compute:
[
P(X = 0) = \frac{e^{-0.5} (0.5)^0}{0!} = e^{-0.5} \approx 0.6065
]
[
P(X = 1) = \frac{e^{-0.5} (0.5)^1}{1!} = 0.6065 \times 0.5 = 0.3033
]
[
P(X = 2) = \frac{e^{-0.5} (0.5)^2}{2!} = 0.6065 \times 0.25 = 0.0758
]
[
P(X = 3) = \frac{e^{-0.5} (0.5)^3}{3!} = 0.6065 \times \frac{0.125}{6} = 0.0126
]
Summing these:
[
P(X \leq 3) = 0.6065 + 0.3033 + 0.0758 + 0.0126 = 0.9982
]
Step 3: Compute ( P(X > 3) )
[
P(X > 3) = 1 – P(X \leq 3) = 1 – 0.9982 = 0.0018
]
Conclusion:
The probability that the insurance company must pay off on more than 3 insured risks in a given year is 0.0018, or 0.18%.