Probability of Union of 3 Sets

Question

Say you know $P(F)$, $P(G)$, $P(H)$, $P(FG)$, $P(FH)$, $P(GH)$ and $P(FGH)$.

How do you find $P(F^CG^CH)$?

$P(F^CG^CH)$

= $P((F^CG^C)H)$ by commutative property.

= $P((F \cup G)^CH)$ by De Morgan's

= ... ?

Answer 1

You were on the right track.

$\begin{align} & P(F^C\cap G^C\cap H) \\ &= P((F^C\cap G^C)\cap H)&& \text{by associative property.} \\ &= P((F \cup G)^C\cap H) &&\text{by De Morgan's} \\[2ex] &= P(H)-P((F\cup G)\cap H) && \text{by relative complementation} \\ &=P(H)-P((F\cap H)\cup(G\cap H)) &&\text{by distribution} \\ &=P(H)-P(F\cap H)-P(G\cap H)+P((F\cap H)\cap(G\cap H)) && \text{by PIE (Principle of Inclusion and Exclusion)} \\ &=P(H)-P(F\cap H)-P(G\cap H)+P(F\cap G\cap H) && \text{by Distribution} \\ \end{align}$

Answer 2

Observe that $$ P(F^c\cap G^c\cap H)=P(H)-P(F\cap H)-P(G\cap H)+P(F\cap G\cap H). $$ Subtract the things that are in $F\cap H$ and $G\cap H$ but then we subtracted the things in $F\cap G\cap H$ twice so we add it back.

Answer 3

Found the answer.

Theorems used in this solution:

$P(A^C) + P(A) = P(Ω)$

$P(A \cup B) = P(A) + P(B) - P(AB)$