This is a homework question that I'm stuck on and I'm looking to see if I'm going about it the right way and how to put the pieces together. So far I know that I can get $P(A \cup B)$ which is $0.15 + 0.10 - 0.04 = 0.21$ and I know that I can get $P(B^c)$ which is $1 - 0.10 = 0.90$, and I know I can get $P(A \cup B')$ by doing $1 - 0.21$ which is $0.79$. However I'm not sure what rule's I can use or how to use these individual parts to get $P(A \cup B^c)$????
Edited: This should have read $P(A \cup B)'$, what's needed is $P(A \cup B^c)$.
$\endgroup$ 21 Answer
$\begingroup$$P(A \cap B^c) = P(A) - P(A \cap B)$ (how?)
Once this is settled, rest follows easily.
$P(A \cup B^c)=P(A)+P(B^c)-P(A \cap B^C)$
$= P(A)+P(B^c) - P(A) + P(A \cap B)$
$ = P(B^c)+ P(A \cap B)$ $ = 0.90 + 0.04 = 0.94$
As you rightly note in the comments, there are multiple ways of reaching this result.
$P(A \cup B^c)=P(A)+P(B^c)-P(A \cap B^C) = P(A)+P(B^c \cap A^c) = P(A)+ 1 - P(B \cup A) $
because $P(B^c)-P(A \cap B^C) = P(A^c \cap B^c) $
$\endgroup$ 3
