Let A, B, and C be events such that [math]\operatorname{P}[A] = 0.2[/math], [math]\operatorname{P}[B] = 0.1 [/math], and [math]\operatorname{P}[C] = 0.3 [/math]. The events A and B are independent, the events B and C are independent, and the events A and C are mutually exclusive.

Calculate [math]\operatorname{P}[A \cup B \cup C] . [/math]

• 0.496
• 0.540
• 0.544
• 0.550
• 0.600

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

Solution: D

[[math]] \begin{align*} \operatorname{P}[ A ∪ B ∪ C ] &= \operatorname{P}[ A] + \operatorname{P}[ B] + \operatorname{P}[C ] − \operatorname{P}[ A ∩ B] − \operatorname{P}[ A ∩ C ] − \operatorname{P}[ B ∩ C ] + \operatorname{P}[ A ∩ B ∩ C ]\\ &= 0.2 + 0.1 + 0.3 − 0.2(0.1) − 0 − 0.1(0.3) + 0 \\ &= 0.55. \end{align*} [[/math]]

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