We have $$P(A) = 0.3$$, $$P(B) = 0.4$$, $$P(C)=0.5$$. We know that the events are mutually independent. We are looking for $$P( \overline{\rm A} \cap \overline{\rm B} \cap C) =?$$ My guess is $$0.7 \cdot 0.6 \cdot 0.5 = 0.21$$. But this wasn't the answer. Any tips?

• Why do you say it is not correct? – Satish Ramanathan Feb 1 '19 at 14:20
• it wasn't on the answer list. – Simon Jachson Feb 1 '19 at 14:20
• What is the answer that you are comparing it with? – Satish Ramanathan Feb 1 '19 at 14:20
• I don't know the correct answer. – Simon Jachson Feb 1 '19 at 14:21
• What were on the answer list? – Satish Ramanathan Feb 1 '19 at 14:22

If $$A$$, $$B$$ and $$C$$ are independent, then $$\overline{\rm A}$$, $$\overline{\rm B}$$ and $$C$$ are independent.
$$P( \overline{\rm A} \cap \overline{\rm B} \cap C) = P( (\overline{\rm {A \cup B}}) \cap C)$$ $$=P( A \cup B \cup C) - P({A \cup B})$$ $$=P(C) - P(B \cap C) - P({A \cap C}) + P( A \cap B \cap C)$$ $$=P(C)-P(A).P(C)-P(B).P(C)+P(A).P(B).P(C)$$ $$=(1-P(A)).(1-P(B)).P(C)$$ $$=P(\overline {\rm A}).P(\overline {\rm B}).P(C)$$