A life insurance company has found there is a 3% probability that a randomly selected application contains an error. Assume applications are mutually independent in this respect. An auditor randomly selects 100 applications.

Calculate the probability that 95% or less of the selected applications are error-free.

• 0.08
• 0.10
• 0.13
• 0.15
• 0.18

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

Solution: E

The number that have errors is a binomial random variable with p = 0.03 and n = 100. Let X be the number that have errors. Then,

[[math]] \begin{align*} \operatorname{P}(\textrm{number that are error-free} ≤ 95) &=\operatorname{P}( X ≥ 5) =1 − \operatorname{P}(0) − \operatorname{P}(1) − \operatorname{P}(2) − \operatorname{P}(3) − \operatorname{P}(4) \\ &= 1- \binom{100}{0}(0.03)^0(0.97)^{100} - \binom{100}{1}(0.03)(0.97)^{99} - \binom{100}{2}(0.03)^2(0.97)^{98}\\ &- \binom{100}{3}(0.03)^3(0.97)^{97} - \binom{100}{4}(0.03)^4(0.97)^{96} \\ &= 0.1821. \end{align*} [[/math]]

Or, the Poisson approximation can be used. Then, [math]\lambda = 3 [/math] and

[[math]] \begin{align*} \operatorname{P}(X \geq 5) &= 1 - \frac{e^{-3}3^0}{0!} - \frac{e^{-3}3}{1} - \frac{e^{-3}e^2}{2!} - \frac{e^{-3}e^3}{3!} - \frac{e^{-3}3^4}{4!} \\ &= 1-e^{-3}(1 +3 + \frac{9}{2} + \frac{27}{6} + \frac{81}{24}) \\ &= 0.1847. \end{align*} [[/math]]

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