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?
$\begingroup$
$\endgroup$
-
$\begingroup$ Why do you say it is not correct? $\endgroup$ – Satish Ramanathan Feb 1 '19 at 14:20
-
$\begingroup$ it wasn't on the answer list. $\endgroup$ – Simon Jachson Feb 1 '19 at 14:20
-
$\begingroup$ What is the answer that you are comparing it with? $\endgroup$ – Satish Ramanathan Feb 1 '19 at 14:20
-
$\begingroup$ I don't know the correct answer. $\endgroup$ – Simon Jachson Feb 1 '19 at 14:21
-
$\begingroup$ What were on the answer list? $\endgroup$ – Satish Ramanathan Feb 1 '19 at 14:22
$\begingroup$
$\endgroup$
0.21 is the correct answer.
If $A$, $B$ and $C$ are independent, then $\overline{\rm A}$, $\overline{\rm B}$ and $C$ are independent.
$P( \overline{\rm A} \cap \overline{\rm B} \cap C) = P( (\overline{\rm {A \cup B}}) \cap C)$ $=P( A \cup B \cup C) - P({A \cup B})$ $=P(C) - P(B \cap C) - P({A \cap C}) + P( A \cap B \cap C)$ $=P(C)-P(A).P(C)-P(B).P(C)+P(A).P(B).P(C)$ $=(1-P(A)).(1-P(B)).P(C)$ $=P(\overline {\rm A}).P(\overline {\rm B}).P(C)$
