What is the probability that four (4) of these 10 insurance policies will have been surrendered before their maturity date?

Steps to Solve

1. Identify the parameters of the problem

We are dealing with a binomial probability situation where we have:

• Number of trials ( n = 10 ) (the total number of policies)
• Number of successes ( k = 4 ) (policies surrendered)
• Probability of success ( p = 0.2 ) (probability that a policy is surrendered)

2. Use the binomial probability formula

The binomial probability formula is given by:

$$ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} $$

Where ( \binom{n}{k} ) is the binomial coefficient.

3. Calculate the binomial coefficient

Calculate ( \binom{10}{4} ):

$$ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 $$

4. Calculate each component of the formula

Now, substitute the values into the formula:

First, calculate ( p^k ):

$$ p^k = (0.2)^4 = 0.0016 $$

Now calculate ( (1 - p)^{n-k} ):

$$ (1 - p)^{n-k} = (0.8)^{10 - 4} = (0.8)^6 = 0.262144 $$

5. Combine all components to find the probability

Now put everything together:

$$ P(X = 4) = 210 \times 0.0016 \times 0.262144 $$

Calculate this:

$$ P(X = 4) = 210 \times 0.0016 \times 0.262144 \approx 0.088 \text{ (rounded to three decimal places)} $$