An insurance company studies back injury claims from a manufacturing company. The insurance company finds that 40% of workers do no lifting on the job, 50% do moderate lifting and 10% do heavy lifting.

During a given year, the probability of filing a claim is 0.05 for a worker who does no lifting, 0.08 for a worker who does moderate lifting and 0.20 for a worker who does heavy lifting.

A worker is chosen randomly from among those who have filed a back injury claim. Calculate the probability that the worker’s job involves moderate or heavy lifting.

• 0.75
• 0.81
• 0.85
• 0.86
• 0.89

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

Solution: A

Define the events as follows:

C = files a claim

N = no lifting

M = moderate lifting

H = heavy lifting

Then, using Bayes’ Theorem,

[[math]] \begin{align*} \operatorname{P}[M \cup H | C) = 1- \operatorname{P}[N | C) &= 1 - \frac{\operatorname{P}[C|N)\operatorname{P}[N)}{\operatorname{P}(C | N ) \operatorname{P}( N ) + \operatorname{P}(C | M) \operatorname{P}[M) + \operatorname{P}[C | H) \operatorname{P}[H)} \\ &= 1- \frac{0.05(0.4)}{0.05(0.4) + 0.08(0.5) + 0.2(0.1)} \\ &= 1-0.25 \\ &= 0.75. \end{align*} [[/math]]

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