A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is 0.60. The numbers of accidents that occur in different months are mutually independent.

Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.

• 0.01
• 0.12
• 0.23
• 0.29
• 0.41

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

If a month with one or more accidents is regarded as success and [math]k[/math] = the number of failures before the fourth success, then [math]k[/math] follows a negative binomial distribution and the requested probability is

[[math]] \begin{align*} \operatorname{P}[ k ≥ 4] = 1 − \operatorname{P}[ k ≤ 3] &= 1 - \sum_{k=0}^3 \binom{3+k}{k} (\frac{3}{5})^4(\frac{2}{5})^k \\ &= 1-(\frac{3}{5})^4\left[ \binom{3}{0}(\frac{2}{5})^0 + \binom{4}{1}\left( \frac{2}{5}\right)^1 + \binom{5}{2} \left( \frac{2}{5}\right)^2 + \binom{6}{3} \left ( \frac{2}{5}\right)^3 \right ] \\ &= 1-\left( \frac{3}{5} \right )^4 \left[ 1 + \frac{8}{5} + \frac{8}{5} + \frac{32}{25} \right ] \\ &= 0.2898. \end{align*} [[/math]]

Alternatively, the solution is

[[math]] \left ( \frac{2}{5} \right )^4 + \binom{4}{1}\left ( \frac{2}{5} \right )^4 \frac{3}{5} + \binom{5}{2} \left ( \frac{2}{5} \right )^4 \left ( \frac{3}{5} \right )^2 + \binom{6}{3} \left ( \frac{2}{5} \right )^4\left ( \frac{3}{5} \right )^3 = 0.2898 [[/math]]

which can be derived directly or by regarding the problem as a negative binomial distribution with

1. success taken as a month with no accidents
2. [math]k[/math] = the number of failures before the fourth success, and
3. calculating [math]\operatorname{P}[ k ≤ 3][/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.