Given that $\textrm{P}(Cc) = 0.82, \textrm{P}(B \cup C) = 0.41, \textrm{P}(A \cup C) = 0.52, \textrm{P}(A \cup B \cup C) = 0.69$. The event $C$ is disjoint from events $A$ and $B$. Compute
a.) $\textrm{P}(Ac ∪ Bc ∪ C)$
b. $\textrm{P}(Ac ∩B ∩ Cc).$
Here is what I have worked out:
a.)
$\textrm{P}(C) = 1 - \textrm{P}(Cc)$
$\textrm{P}(C) = 1 - 0.82$
$\textrm{P}(C) = 0.18$
$\textrm{P}(A ∪ C) = 0.52$
$\textrm{P}(A) + \textrm{P}(C) - 0 = 0.52$
$\textrm{P}(A) = 0.52 - 0.18$
$\textrm{P}(A) = 0.34, \textrm{P}(Ac) = 0.66$
$\textrm{P}(B ∪ C) = 0.41$
$\textrm{P}(B) + \textrm{P}(C) - 0 = 0.41$
$\textrm{P}(B) = 0.41 - 0.18$
$\textrm{P}(B) = 0.23, \textrm{P}(Bc) = 0.77$
$\textrm{P}(Ac ∪ Bc ∪ C) = \textrm{P}(Ac) + \textrm{P}(Bc) + \textrm{P}(C) - \textrm{P}(A ∪ C) - \textrm{P}(B ∪ C)$
$\textrm{P}(Ac ∪ Bc ∪ C) = 0.66 + 0.77 + 0.18 - 0.52 - 0.41$
$\textrm{P}(Ac ∪ Bc ∪ C) = 0.68$
b.
$\textrm{P} (Ac ∩B ∩ Cc) = 0$
Since $C$ is disjoint from $A$ and $B$, they have no intersections.
I am having a hard time trying to visualize how to calculate such a problem. Is my line of thought correct and if not, could it be explained where I went wrong. (if possible with a venn diagram on top of the calculation?)
From $\Pr(A\cup B\cup C)=0.69$, from $\Pr(C)=1-\Pr(C^c)=1-0.82=0.18$, and from $C$ is disjoint from $A$ and from $B$ (and thus disjoint from $A\cup B$) we get:
$$\Pr(A\cup B) = \Pr(A\cup B\cup C)-\Pr(C)=0.68-0.18=0.5$$
We similarly find $\Pr(A)=\Pr(A\cup C)-\Pr(C)=0.52-0.18=0.34$ and $\Pr(B)=\Pr(B\cup C)-\Pr(C)=0.41-0.18=0.23$
By a rearrangement of inclusion-exclusion, $\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)=0.34+0.23-0.5=0.07$
We get finally then looking at the venn diagram or by standard results $\Pr(A^c\cup B^c\cup C)=\Pr(A^c\cup B^c) = 1-\Pr(A\cap B)=0.93$
Your answer for part (b) was correct.