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Q.
If $P ( E )=0.15, P ( F )=0.09$ and $P ( E \cup F )=0.19$, then the value of $P \left( E ^{ C } \cap F \right)$ is
Probability - Part 2
Solution:
$P \left( E ^{ C } \cap F \right)= P ( F )- P ( E \cap F ) \ldots(1)$
$\text { and } P(E \cap F)=P(E)+P(F)-P(E \cup F) $
$ P(E \cap F)=0.15+0.09-0.19=0.05$
$\therefore P\left(E^C \cap F\right)=0.09-0.05=0.04$